Appendix A — Essential Real Analysis for Finite Element Methods

Purpose. This appendix supplements Chapter 2 and reviews a small collection of real-analysis concepts that appear repeatedly in finite element analysis and Sobolev-space arguments. The emphasis is on geometric and analytical meaning, useful examples, and the distinctions that are easy to blur: maximum versus supremum, closed versus bounded, pointwise versus uniform convergence, and supremum versus essential supremum.

A.1 Sets in $\mathbb{R}^d$: balls, neighborhoods, interior, closure, and boundary

For $\bm{x}\in\mathbb{R}^d$ and $r>0$, the open ball centered at $\bm{x}$ with radius $r$ is \[B_r(\bm{x})=\{\bm{y}\in\mathbb{R}^d:\left\lVert \bm{y}-\bm{x}\right\rVert_2<r\}.\] The corresponding closed ball is \[\overline B_r(\bm{x})=\{\bm{y}\in\mathbb{R}^d:\left\lVert \bm{y}-\bm{x}\right\rVert_2\le r\}.\] A set $U\subset\mathbb{R}^d$ is open if every point of $U$ has some open ball contained entirely in $U$: \[\forall \bm{x}\in U,\qquad \exists r>0\quad\text{such that}\quad B_r(\bm{x})\subset U.\] A set $F\subset\mathbb{R}^d$ is closed if its complement $\mathbb{R}^d\setminus F$ is open.

The interior of a set $A$, denoted $A^\circ$, is the union of all open subsets of $A$. Equivalently, $\bm{x}\in A^\circ$ if some open ball centered at $\bm{x}$ lies in $A$.

The closure $\overline A$ consists of $A$ together with all points that can be approached arbitrarily closely by points of $A$. Equivalently, \[\bm{x}\in\overline A \quad\Longleftrightarrow\quad B_r(\bm{x})\cap A\ne\varnothing\quad\text{for every }r>0.\] The boundary is \[\partial A=\overline A\setminus A^\circ.\] A point on $\partial A$ has points of $A$ and points outside $A$ arbitrarily close to it.

A useful sequential characterization is:

Closed-set criterion. A set $F\subset\mathbb{R}^d$ is closed if and only if every convergent sequence $\{\bm{x}_n\}\subset F$ has its limit in $F$. Thus \[\bm{x}_n\in F,\quad \bm{x}_n\to\bm{x} \quad\Longrightarrow\quad \bm{x}\in F.\] This criterion is often easier to use than the complement-based definition of closedness.

Example A.1 (Open, closed, and neither) In $\mathbb{R}$, the interval $(0,1)$ is open, $[0,1]$ is closed, and $[0,1)$ is neither open nor closed. The set $\mathbb{R}$ and the empty set are both open and closed. In $\mathbb{R}^d$, an open ball $B_r(\bm{x})$ is open and a closed ball $\overline B_r(\bm{x})$ is closed.

Why this matters for PDEs

A physical domain $\Omega$ is usually modeled as an open set because differential equations are imposed at interior points. Boundary conditions are prescribed on $\partial\Omega$. Many analytical results require additional regularity of the boundary, such as the Lipschitz assumption used in Chapter 2.

A.2 Bounded sets and compact sets

A set $A\subset\mathbb{R}^d$ is bounded if it fits inside some sufficiently large ball: there are $\bm{x}_0\in\mathbb{R}^d$ and $R>0$ such that \[A\subset B_R(\bm{x}_0).\] Boundedness and closedness are different properties. The interval $(0,1)$ is bounded but not closed. The set $[0,\infty)$ is closed but unbounded.

A set $K\subset\mathbb{R}^d$ is compact if every open cover of $K$ has a finite subcover. For work in finite-dimensional Euclidean spaces, the following characterization is usually the most useful.

Heine–Borel theorem. A subset $K\subset\mathbb{R}^d$ is compact if and only if it is closed and bounded.

An equally useful sequential formulation is:

Sequential compactness. A set $K\subset\mathbb{R}^d$ is compact if and only if every sequence in $K$ has a subsequence that converges to a point in $K$.

Example A.2 (A bounded set that is not compact) The interval $(0,1)$ is bounded, but it is not compact because it is not closed. The sequence \[x_n=\frac1n\] lies in $(0,1)$ and converges to $0$, which does not belong to the interval.

Compactness becomes much subtler in infinite-dimensional function spaces. Closed and bounded subsets of an infinite-dimensional normed space need not be compact. This distinction becomes important in advanced existence and approximation theory, but the finite-dimensional Heine–Borel theorem is enough for the present supplement.

A.3 Sequences, limits, subsequences, and Cauchy sequences

A sequence in $\mathbb{R}^d$ is an ordered list $\{\bm{x}_n\}_{n=1}^\infty$. We say \[\bm{x}_n\to\bm{x}\] if for every $\varepsilon>0$ there exists $N$ such that \[n\ge N\quad\Longrightarrow\quad \left\lVert \bm{x}_n-\bm{x}\right\rVert_2<\varepsilon.\] The limit, if it exists, is unique.

A subsequence is formed by selecting indices \[n_1<n_2<n_3<\cdots\] and considering $\{\bm{x}_{n_k}\}$. A convergent sequence and all of its subsequences have the same limit. The converse is false: a nonconvergent sequence can have convergent subsequences.

A sequence is Cauchy if its terms become arbitrarily close to one another: \[\forall\varepsilon>0,\quad \exists N\quad\text{such that}\quad m,n\ge N\Longrightarrow \left\lVert \bm{x}_m-\bm{x}_n\right\rVert_2<\varepsilon.\] Every convergent sequence is Cauchy. In $\mathbb{R}^d$, every Cauchy sequence converges. This completeness property is one of the prototypes for Banach and Hilbert spaces.

Bolzano–Weierstrass theorem. Every bounded sequence in $\mathbb{R}^d$ has a convergent subsequence.

Example A.3 (A sequence with two convergent subsequences) The sequence $x_n=(-1)^n$ does not converge. Its even subsequence converges to $1$, while its odd subsequence converges to $-1$.

A.4 Upper and lower bounds, supremum and infimum

Let $A\subset\mathbb{R}$ be nonempty. A number $M\in\mathbb{R}$ is an upper bound for $A$ if \[a\le M\qquad\forall a\in A.\] A number $m$ is a lower bound if \[m\le a\qquad\forall a\in A.\] If $A$ has an upper bound, it is bounded above. If it has a lower bound, it is bounded below.

The supremum $\sup A$ is the least upper bound of $A$. Thus $s=\sup A$ means

$a\le s$ for all $a\in A$;
for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a$.

Similarly, $i=\inf A$ is the greatest lower bound.

A maximum is an element of the set. We write $M=\max A$ if \[M\in A,\qquad a\le M\quad\forall a\in A.\] Therefore a maximum is always a supremum, but a supremum need not be a maximum. Likewise, a minimum is always an infimum, but an infimum need not be a minimum.

Example A.4 (Supremum without a maximum) For $A=(0,1)$, \[\sup A=1,\qquad \inf A=0,\] but $A$ has neither a maximum nor a minimum because neither endpoint belongs to the set.

Example A.5 (Supremum attained) For $A=[0,1]$, \[\max A=\sup A=1, \qquad \min A=\inf A=0.\] The difference is not the numerical values. It is whether the extremal values are attained by elements of the set.

The completeness of $\mathbb{R}$ can be expressed through the least-upper-bound property: every nonempty subset of $\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$.

A.5 Extrema of functions

For a real-valued function $f:A\to\mathbb{R}$, define \[\sup_A f=\sup\{f(x):x\in A\}, \qquad \inf_A f=\inf\{f(x):x\in A\}.\] A maximum is attained at some $x_*\in A$ if \[f(x_*)\ge f(x)\qquad\forall x\in A.\] Then $f(x_*)=\max_A f=\sup_A f$.

Example A.6 (The domain controls whether an extremum exists) Let $f(x)=x$.

On $(0,1)$, $\sup f=1$ and $\inf f=0$, but neither is attained.
On $[0,1]$, $\max f=1$ and $\min f=0$.

The formula for the function has not changed. Only the domain has changed.

A.5.1 Local and global extrema

A point $\bm{x}_*\in A$ is a global maximum point if \[f(\bm{x}_*)\ge f(\bm{x})\qquad\forall\bm{x}\in A.\] It is a local maximum point if the same inequality holds only for points of $A$ sufficiently close to $\bm{x}_*$. Local minima are defined analogously. A global extremum is automatically local, but a local extremum need not be global.

If $f$ is differentiable on an open set and has a local extremum at an interior point $\bm{x}_*$, then Fermat’s necessary condition gives \[\nabla f(\bm{x}_*)=\bm{0}.\] The converse is false: a stationary point need not be a maximum or minimum. This distinction becomes important in energy minimization and optimization.

A.6 Continuity

Let $A\subset\mathbb{R}^d$ and $f:A\to\mathbb{R}^m$. The function is continuous at $\bm{x}\in A$ if for every $\varepsilon>0$ there exists $\delta>0$ such that \[\bm{y}\in A,\quad \left\lVert \bm{y}-\bm{x}\right\rVert_2<\delta \quad\Longrightarrow\quad \left\lVert f(\bm{y})-f(\bm{x})\right\rVert_2<\varepsilon.\] It is continuous on $A$ if it is continuous at every point of $A$.

A particularly useful equivalent statement is the sequential characterization: \[\bm{x}_n\to\bm{x}\text{ in }A \quad\Longrightarrow\quad f(\bm{x}_n)\to f(\bm{x}).\]

Continuity is defined relative to the domain. A boundary point of a closed interval, for example, is tested only with points that remain in the interval. Thus $f:[0,1]\to\mathbb{R}$ can be continuous at $0$ even though the domain contains no points to the left of $0$.

A.6.1 Continuity on open and closed sets

There is nothing intrinsically stronger about “continuity on a closed set” than “continuity on an open set.” The important difference is that compact domains give additional consequences.

Continuous image of a compact set. If $K\subset\mathbb{R}^d$ is compact and $f:K\to\mathbb{R}^m$ is continuous, then $f(K)$ is compact. In particular, a real-valued continuous function on a compact set has a compact range.

Extreme value theorem. If $K\subset\mathbb{R}^d$ is compact and $f:K\to\mathbb{R}$ is continuous, then $f$ is bounded and attains both a maximum and a minimum: \[\exists \bm{x}_{\min},\bm{x}_{\max}\in K \quad\text{such that}\quad f(\bm{x}_{\min})\le f(\bm{x})\le f(\bm{x}_{\max}) \quad\forall \bm{x}\in K.\]

Heine–Cantor theorem. If $K\subset\mathbb{R}^d$ is compact and $f:K\to\mathbb{R}^m$ is continuous, then $f$ is uniformly continuous on $K$.

Example A.7 (Continuity alone does not guarantee a maximum) The function $f(x)=x$ is continuous on the open interval $(0,1)$, but it has no maximum there. Compactness of the domain is what allows the extreme value theorem to apply.

A.7 Uniform continuity and Lipschitz continuity

A function $f:A\to\mathbb{R}^m$ is uniformly continuous if \[\forall\varepsilon>0,\quad \exists\delta>0\quad\text{such that}\quad \left\lVert \bm{x}-\bm{y}\right\rVert_2<\delta \Longrightarrow \left\lVert f(\bm{x})-f(\bm{y})\right\rVert_2<\varepsilon\] for all $\bm{x},\bm{y}\in A$. The key point is that one value of $\delta$ works everywhere in $A$.

A function is Lipschitz continuous if there exists $L\ge0$ such that \[\left\lVert f(\bm{x})-f(\bm{y})\right\rVert_2 \le L\left\lVert \bm{x}-\bm{y}\right\rVert_2 \qquad\forall\bm{x},\bm{y}\in A.\] Every Lipschitz function is uniformly continuous, and every uniformly continuous function is continuous. The converses are false in general.

Example A.8 (Continuous but not uniformly continuous) The function $f(x)=1/x$ is continuous on $(0,1)$ but not uniformly continuous there. Its variation becomes arbitrarily rapid near $x=0$.

Lipschitz regularity appears in FEM in several ways: geometric mappings, Lipschitz domains, nonlinear constitutive mappings, and estimates in which bounded derivatives imply controlled changes in a function.

A.7.1 A useful derivative-to-Lipschitz estimate

In one dimension, the mean value theorem implies that if $f$ is differentiable on an interval and $|f'(x)|\le L$, then \[|f(x)-f(y)|\le L|x-y|.\] More generally, if $A\subset\mathbb{R}^d$ is convex, $f:A\to\mathbb{R}$ is differentiable, and $\left\lVert \nabla f(\bm{x})\right\rVert_2\le L$ throughout $A$, then \[|f(\bm{x})-f(\bm{y})|\le L\left\lVert \bm{x}-\bm{y}\right\rVert_2.\] This is one common route from derivative bounds to continuity estimates.

A.8 Almost everywhere, essential supremum, and essential infimum

Lebesgue spaces identify functions that differ only on sets of measure zero. This motivates the essential supremum and essential infimum.

A property holds almost everywhere (a.e.) on $\Omega$ if it fails only on a set of measure zero. In $\mathbb{R}^d$, individual points and finite or countable sets have measure zero. A curve in a three-dimensional domain also has three-dimensional measure zero.

For a measurable function $f:\Omega\to\mathbb{R}$, the essential supremum is \[\operatorname*{ess\,sup}_{x\in\Omega} f(x) = \inf\{M\in\mathbb{R}:f(x)\le M\text{ for almost every }x\in\Omega\}.\] The essential infimum is \[\operatorname*{ess\,inf}_{x\in\Omega} f(x) = \sup\{m\in\mathbb{R}:m\le f(x)\text{ for almost every }x\in\Omega\}.\] The adjective essential means that exceptional values on measure-zero sets are ignored.

Example A.9 (Supremum and essential supremum can differ) Define $f:[0,1]\to\mathbb{R}$ by \[f(x)= \begin{cases} 100, & x=1/2,\\ 1, & x\ne 1/2. \end{cases}\] Then \[\sup_{[0,1]}f=100, \qquad \operatorname*{ess\,sup}_{[0,1]}f=1.\] The single exceptional point has measure zero.

The $L^\infty$ norm is defined through the essential supremum: \[\left\lVert f\right\rVert_{L^\infty(\Omega)} = \operatorname*{ess\,sup}_{x\in\Omega}|f(x)|.\] This makes the norm insensitive to changes on measure-zero sets, consistent with the way $L^p$ spaces are defined.

For a coefficient field $k(\bm{x})$ in an elliptic PDE, assumptions such as \[0<k_{\min} \le k(\bm{x}) \le k_{\max}<\infty \qquad\text{a.e. in }\Omega\] are more naturally expressed as \[0<\operatorname*{ess\,inf}_{\Omega}k, \qquad \operatorname*{ess\,sup}_{\Omega}k<\infty.\] These bounds later enter coercivity and continuity estimates.

A.9 Sequences of functions: pointwise and uniform convergence

Let $f_n:A\to\mathbb{R}$ be a sequence of functions.

The sequence converges pointwise to $f$ if, for every fixed $x\in A$, \[f_n(x)\to f(x).\] The index $N$ required to make $|f_n(x)-f(x)|<\varepsilon$ is allowed to depend on $x$.

The sequence converges uniformly to $f$ if \[\sup_{x\in A}|f_n(x)-f(x)|\to0.\] Equivalently, for every $\varepsilon>0$ there exists one $N$ that works for all $x\in A$.

Uniform convergence is stronger than pointwise convergence. A uniform limit of continuous functions is continuous. A pointwise limit of continuous functions need not be continuous.

Example A.10 (Pointwise but not uniform convergence) On $[0,1]$, let \[f_n(x)=x^n.\] Then \[f_n(x)\to \begin{cases} 0, & 0\le x<1,\\ 1, & x=1. \end{cases}\] The limit is discontinuous, so the convergence cannot be uniform.

In finite element analysis, one often studies convergence in a norm such as $L^2$ or $H^1$, which is different from both pointwise and uniform convergence. The important habit is to always ask: in what sense is the approximation converging?

A.10 A compact list of implications worth remembering

For subsets of $\mathbb{R}^d$ and functions defined on them: \[\boxed{\text{compact in }\mathbb{R}^d\iff\text{closed and bounded}}\] \[\boxed{\text{continuous on compact}\Longrightarrow\text{bounded and attains max/min}}\] \[\boxed{\text{continuous on compact}\Longrightarrow\text{uniformly continuous}}\] \[\boxed{\text{Lipschitz}\Longrightarrow\text{uniformly continuous}\Longrightarrow\text{continuous}}\] \[\boxed{\max A\text{ exists}\Longrightarrow\max A=\sup A,\quad \min A\text{ exists}\Longrightarrow\min A=\inf A}\] No reverse implication in these statements should be assumed without additional hypotheses.

A.11 Practice problems

Exercise A.1 (A. Sets and compactness)

Classify each set as open, closed, both, or neither in $\mathbb{R}$: $(0,2)$, $[0,2]$, $[0,2)$, $\mathbb{R}$, $\varnothing$.
For $A=(0,1]\subset\mathbb{R}$, determine $A^\circ$, $\overline A$, and $\partial A$.
Is the set $\{1/n:n\in\mathbb N,\ n\ge1\}$ closed in $\mathbb{R}$? Is it compact? Explain.
Determine whether each set is compact in $\mathbb{R}^2$: the open unit disk, the closed unit disk, the line segment $\{(x,0):0\le x\le1\}$, and the whole $x$-axis.

Exercise A.2 (B. Sequences and limits)

Determine the limit of $x_n=(1/n,\,1-1/n)$ in $\mathbb{R}^2$.
Show that $x_n=(-1)^n+1/n$ does not converge, identify two convergent subsequences, and state the limit of each subsequence.
Determine whether $x_n=\log(n+1)-\log n$ is Cauchy.
Give an example of a bounded sequence that does not converge but has a convergent subsequence.

Exercise A.3 (C. Supremum, infimum, maximum, minimum) When a requested maximum or minimum does not exist, state why it does not exist.

For each set, determine $\sup$, $\inf$, $\max$, and $\min$ when they exist: \[(2,5),\qquad [2,5),\qquad \{1-1/n:n\ge1\},\qquad \{(-1)^n:n\ge1\}.\]
Let $f(x)=x(1-x)$ on $(0,1)$ and on $[0,1]$. Determine the supremum, infimum, maximum, and minimum in each case.
Let $f(x)=1/x$ on $(0,1)$. Is $f$ bounded above? Does it have a minimum?

Exercise A.4 (D. Continuity and extrema)

Explain why $f(x)=x^2+1$ has a maximum and minimum on $[-2,3]$.
Does $f(x)=x^2+1$ have a maximum on $\mathbb{R}$? Does it have a minimum?
Give an example of a continuous function on a bounded open set that does not attain its supremum.
Show that $f(x)=x^2$ is Lipschitz on $[-1,1]$. Find one admissible Lipschitz constant.
Is $f(x)=\sqrt{x}$ Lipschitz on $[0,1]$? Is it uniformly continuous there?

Exercise A.5 (E. Essential bounds)

Define $f:[0,1]\to\mathbb{R}$ by $f(0)=7$ and $f(x)=2$ for $x>0$. Find $\sup f$, $\inf f$, $\operatorname*{ess\,sup}f$, and $\operatorname*{ess\,inf}f$.
Define $f(x)=1$ if $x$ is rational and $f(x)=0$ if $x$ is irrational on $[0,1]$. Determine $\sup f$, $\inf f$, $\operatorname*{ess\,sup}f$, and $\operatorname*{ess\,inf}f$.
Suppose $k(x)=2+x$ for almost every $x\in(0,1)$ but $k(1/2)=1000$. Find $\operatorname*{ess\,inf}k$ and $\operatorname*{ess\,sup}k$.

Exercise A.6 (F. Pointwise and uniform convergence)

For $f_n(x)=x/n$ on $[0,1]$, determine the pointwise limit and show that the convergence is uniform.
For $f_n(x)=x^n$ on $[0,1]$, compute $\sup_{x\in[0,1]}|f_n(x)-f(x)|$ for the pointwise limit $f$ and explain why convergence is not uniform.
Let $f_n(x)=1/(n+x)$ on $[0,\infty)$. Determine whether $f_n\to0$ uniformly.

Selected answers

A. Sets and compactness

$(0,2)$ open; $[0,2]$ closed; $[0,2)$ neither; $\mathbb{R}$ and $\varnothing$ both open and closed.
$A^\circ=(0,1)$, $\overline A=[0,1]$, $\partial A=\{0,1\}$.
The set is not closed because $0$ is a limit point not contained in it; therefore it is not compact.
Only the closed unit disk and the line segment are compact.

C. Supremum, infimum, maximum, minimum

$(2,5)$: $\sup=5$, $\inf=2$, no max/min. $[2,5)$: $\sup=5$, $\inf=\min=2$, no max. For $\{1-1/n\}$: $\sup=1$, no max, $\inf=\min=0$. For $\{(-1)^n\}$: $\sup=\max=1$, $\inf=\min=-1$.
On both domains, $\sup=\max=1/4$ at $x=1/2$. On $(0,1)$, $\inf=0$ but no minimum; on $[0,1]$, $\inf=\min=0$.
Unbounded above; minimum $1$ is not attained on $(0,1)$, so there is no minimum and $\inf=1$.

E. Essential bounds

$\sup=7$, $\inf=2$, $\operatorname*{ess\,sup}=\operatorname*{ess\,inf}=2$.
$\sup=1$, $\inf=0$, $\operatorname*{ess\,sup}=0$, $\operatorname*{ess\,inf}=0$ because the rationals have measure zero.
$\operatorname*{ess\,inf}k=2$ and $\operatorname*{ess\,sup}k=3$.

# Essential Real Analysis for Finite Element Methods {#sec-appendix-real-analysis} ```{=latex} \chaptershorttitle{Real Analysis Essentials} ``` **Purpose.** This appendix supplements Chapter 2 and reviews a small collection of real-analysis concepts that appear repeatedly in finite element analysis and Sobolev-space arguments. The emphasis is on geometric and analytical meaning, useful examples, and the distinctions that are easy to blur: maximum versus supremum, closed versus bounded, pointwise versus uniform convergence, and supremum versus essential supremum. ## Sets in $\mathbb{R}^d$: balls, neighborhoods, interior, closure, and boundary For $\bm{x}\in\mathbb{R}^d$ and $r>0$, the **open ball** centered at $\bm{x}$ with radius $r$ is $$B_r(\bm{x})=\{\bm{y}\in\mathbb{R}^d:\left\lVert \bm{y}-\bm{x}\right\rVert_2<r\}.$$ The corresponding **closed ball** is $$\overline B_r(\bm{x})=\{\bm{y}\in\mathbb{R}^d:\left\lVert \bm{y}-\bm{x}\right\rVert_2\le r\}.$$ A set $U\subset\mathbb{R}^d$ is **open** if every point of $U$ has some open ball contained entirely in $U$: $$\forall \bm{x}\in U,\qquad \exists r>0\quad\text{such that}\quad B_r(\bm{x})\subset U.$$ A set $F\subset\mathbb{R}^d$ is **closed** if its complement $\mathbb{R}^d\setminus F$ is open. The **interior** of a set $A$, denoted $A^\circ$, is the union of all open subsets of $A$. Equivalently, $\bm{x}\in A^\circ$ if some open ball centered at $\bm{x}$ lies in $A$. The **closure** $\overline A$ consists of $A$ together with all points that can be approached arbitrarily closely by points of $A$. Equivalently, $$\bm{x}\in\overline A \quad\Longleftrightarrow\quad B_r(\bm{x})\cap A\ne\varnothing\quad\text{for every }r>0.$$ The **boundary** is $$\partial A=\overline A\setminus A^\circ.$$ A point on $\partial A$ has points of $A$ and points outside $A$ arbitrarily close to it. A useful sequential characterization is: **Closed-set criterion.** A set $F\subset\mathbb{R}^d$ is closed if and only if every convergent sequence $\{\bm{x}_n\}\subset F$ has its limit in $F$. Thus $$\bm{x}_n\in F,\quad \bm{x}_n\to\bm{x} \quad\Longrightarrow\quad \bm{x}\in F.$$ This criterion is often easier to use than the complement-based definition of closedness. ::: {#exm-appendix-real-open-closed} ## Open, closed, and neither In $\mathbb{R}$, the interval $(0,1)$ is open, $[0,1]$ is closed, and $[0,1)$ is neither open nor closed. The set $\mathbb{R}$ and the empty set are both open and closed. In $\mathbb{R}^d$, an open ball $B_r(\bm{x})$ is open and a closed ball $\overline B_r(\bm{x})$ is closed. ::: ### Why this matters for PDEs {#why-this-matters-for-pdes .unnumbered} A physical domain $\Omega$ is usually modeled as an open set because differential equations are imposed at interior points. Boundary conditions are prescribed on $\partial\Omega$. Many analytical results require additional regularity of the boundary, such as the Lipschitz assumption used in Chapter 2. ## Bounded sets and compact sets A set $A\subset\mathbb{R}^d$ is **bounded** if it fits inside some sufficiently large ball: there are $\bm{x}_0\in\mathbb{R}^d$ and $R>0$ such that $$A\subset B_R(\bm{x}_0).$$ Boundedness and closedness are different properties. The interval $(0,1)$ is bounded but not closed. The set $[0,\infty)$ is closed but unbounded. A set $K\subset\mathbb{R}^d$ is **compact** if every open cover of $K$ has a finite subcover. For work in finite-dimensional Euclidean spaces, the following characterization is usually the most useful. **Heine--Borel theorem.** A subset $K\subset\mathbb{R}^d$ is compact if and only if it is closed and bounded. An equally useful sequential formulation is: **Sequential compactness.** A set $K\subset\mathbb{R}^d$ is compact if and only if every sequence in $K$ has a subsequence that converges to a point in $K$. ::: {#exm-appendix-real-bounded-not-compact} ## A bounded set that is not compact The interval $(0,1)$ is bounded, but it is not compact because it is not closed. The sequence $$x_n=\frac1n$$ lies in $(0,1)$ and converges to $0$, which does not belong to the interval. ::: Compactness becomes much subtler in infinite-dimensional function spaces. Closed and bounded subsets of an infinite-dimensional normed space need not be compact. This distinction becomes important in advanced existence and approximation theory, but the finite-dimensional Heine--Borel theorem is enough for the present supplement. ## Sequences, limits, subsequences, and Cauchy sequences A sequence in $\mathbb{R}^d$ is an ordered list $\{\bm{x}_n\}_{n=1}^\infty$. We say $$\bm{x}_n\to\bm{x}$$ if for every $\varepsilon>0$ there exists $N$ such that $$n\ge N\quad\Longrightarrow\quad \left\lVert \bm{x}_n-\bm{x}\right\rVert_2<\varepsilon.$$ The limit, if it exists, is unique. A **subsequence** is formed by selecting indices $$n_1<n_2<n_3<\cdots$$ and considering $\{\bm{x}_{n_k}\}$. A convergent sequence and all of its subsequences have the same limit. The converse is false: a nonconvergent sequence can have convergent subsequences. A sequence is **Cauchy** if its terms become arbitrarily close to one another: $$\forall\varepsilon>0,\quad \exists N\quad\text{such that}\quad m,n\ge N\Longrightarrow \left\lVert \bm{x}_m-\bm{x}_n\right\rVert_2<\varepsilon.$$ Every convergent sequence is Cauchy. In $\mathbb{R}^d$, every Cauchy sequence converges. This completeness property is one of the prototypes for Banach and Hilbert spaces. **Bolzano--Weierstrass theorem.** Every bounded sequence in $\mathbb{R}^d$ has a convergent subsequence. ::: {#exm-appendix-real-subsequences} ## A sequence with two convergent subsequences The sequence $x_n=(-1)^n$ does not converge. Its even subsequence converges to $1$, while its odd subsequence converges to $-1$. ::: ## Upper and lower bounds, supremum and infimum Let $A\subset\mathbb{R}$ be nonempty. A number $M\in\mathbb{R}$ is an **upper bound** for $A$ if $$a\le M\qquad\forall a\in A.$$ A number $m$ is a **lower bound** if $$m\le a\qquad\forall a\in A.$$ If $A$ has an upper bound, it is **bounded above**. If it has a lower bound, it is **bounded below**. The **supremum** $\sup A$ is the least upper bound of $A$. Thus $s=\sup A$ means 1. $a\le s$ for all $a\in A$; 2. for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a$. Similarly, $i=\inf A$ is the greatest lower bound. A **maximum** is an element of the set. We write $M=\max A$ if $$M\in A,\qquad a\le M\quad\forall a\in A.$$ Therefore a maximum is always a supremum, but a supremum need not be a maximum. Likewise, a minimum is always an infimum, but an infimum need not be a minimum. ::: {#exm-appendix-real-supremum-no-maximum} ## Supremum without a maximum For $A=(0,1)$, $$\sup A=1,\qquad \inf A=0,$$ but $A$ has neither a maximum nor a minimum because neither endpoint belongs to the set. ::: ::: {#exm-appendix-real-supremum-attained} ## Supremum attained For $A=[0,1]$, $$\max A=\sup A=1, \qquad \min A=\inf A=0.$$ The difference is not the numerical values. It is whether the extremal values are attained by elements of the set. ::: The completeness of $\mathbb{R}$ can be expressed through the **least-upper-bound property**: every nonempty subset of $\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$. ## Extrema of functions For a real-valued function $f:A\to\mathbb{R}$, define $$\sup_A f=\sup\{f(x):x\in A\}, \qquad \inf_A f=\inf\{f(x):x\in A\}.$$ A maximum is attained at some $x_*\in A$ if $$f(x_*)\ge f(x)\qquad\forall x\in A.$$ Then $f(x_*)=\max_A f=\sup_A f$. ::: {#exm-appendix-real-domain-extremum} ## The domain controls whether an extremum exists Let $f(x)=x$. - On $(0,1)$, $\sup f=1$ and $\inf f=0$, but neither is attained. - On $[0,1]$, $\max f=1$ and $\min f=0$. The formula for the function has not changed. Only the domain has changed. ::: ### Local and global extrema A point $\bm{x}_*\in A$ is a **global maximum point** if $$f(\bm{x}_*)\ge f(\bm{x})\qquad\forall\bm{x}\in A.$$ It is a **local maximum point** if the same inequality holds only for points of $A$ sufficiently close to $\bm{x}_*$. Local minima are defined analogously. A global extremum is automatically local, but a local extremum need not be global. If $f$ is differentiable on an open set and has a local extremum at an interior point $\bm{x}_*$, then Fermat's necessary condition gives $$\nabla f(\bm{x}_*)=\bm{0}.$$ The converse is false: a stationary point need not be a maximum or minimum. This distinction becomes important in energy minimization and optimization. ## Continuity Let $A\subset\mathbb{R}^d$ and $f:A\to\mathbb{R}^m$. The function is **continuous at** $\bm{x}\in A$ if for every $\varepsilon>0$ there exists $\delta>0$ such that $$\bm{y}\in A,\quad \left\lVert \bm{y}-\bm{x}\right\rVert_2<\delta \quad\Longrightarrow\quad \left\lVert f(\bm{y})-f(\bm{x})\right\rVert_2<\varepsilon.$$ It is continuous on $A$ if it is continuous at every point of $A$. A particularly useful equivalent statement is the sequential characterization: $$\bm{x}_n\to\bm{x}\text{ in }A \quad\Longrightarrow\quad f(\bm{x}_n)\to f(\bm{x}).$$ Continuity is defined relative to the domain. A boundary point of a closed interval, for example, is tested only with points that remain in the interval. Thus $f:[0,1]\to\mathbb{R}$ can be continuous at $0$ even though the domain contains no points to the left of $0$. ### Continuity on open and closed sets There is nothing intrinsically stronger about "continuity on a closed set" than "continuity on an open set." The important difference is that compact domains give additional consequences. **Continuous image of a compact set.** If $K\subset\mathbb{R}^d$ is compact and $f:K\to\mathbb{R}^m$ is continuous, then $f(K)$ is compact. In particular, a real-valued continuous function on a compact set has a compact range. **Extreme value theorem.** If $K\subset\mathbb{R}^d$ is compact and $f:K\to\mathbb{R}$ is continuous, then $f$ is bounded and attains both a maximum and a minimum: $$\exists \bm{x}_{\min},\bm{x}_{\max}\in K \quad\text{such that}\quad f(\bm{x}_{\min})\le f(\bm{x})\le f(\bm{x}_{\max}) \quad\forall \bm{x}\in K.$$ **Heine--Cantor theorem.** If $K\subset\mathbb{R}^d$ is compact and $f:K\to\mathbb{R}^m$ is continuous, then $f$ is uniformly continuous on $K$. ::: {#exm-appendix-real-continuity-no-maximum} ## Continuity alone does not guarantee a maximum The function $f(x)=x$ is continuous on the open interval $(0,1)$, but it has no maximum there. Compactness of the domain is what allows the extreme value theorem to apply. ::: ## Uniform continuity and Lipschitz continuity A function $f:A\to\mathbb{R}^m$ is **uniformly continuous** if $$\forall\varepsilon>0,\quad \exists\delta>0\quad\text{such that}\quad \left\lVert \bm{x}-\bm{y}\right\rVert_2<\delta \Longrightarrow \left\lVert f(\bm{x})-f(\bm{y})\right\rVert_2<\varepsilon$$ for all $\bm{x},\bm{y}\in A$. The key point is that one value of $\delta$ works everywhere in $A$. A function is **Lipschitz continuous** if there exists $L\ge0$ such that $$\left\lVert f(\bm{x})-f(\bm{y})\right\rVert_2 \le L\left\lVert \bm{x}-\bm{y}\right\rVert_2 \qquad\forall\bm{x},\bm{y}\in A.$$ Every Lipschitz function is uniformly continuous, and every uniformly continuous function is continuous. The converses are false in general. ::: {#exm-appendix-real-not-uniformly-continuous} ## Continuous but not uniformly continuous The function $f(x)=1/x$ is continuous on $(0,1)$ but not uniformly continuous there. Its variation becomes arbitrarily rapid near $x=0$. ::: Lipschitz regularity appears in FEM in several ways: geometric mappings, Lipschitz domains, nonlinear constitutive mappings, and estimates in which bounded derivatives imply controlled changes in a function. ### A useful derivative-to-Lipschitz estimate In one dimension, the mean value theorem implies that if $f$ is differentiable on an interval and $|f'(x)|\le L$, then $$|f(x)-f(y)|\le L|x-y|.$$ More generally, if $A\subset\mathbb{R}^d$ is convex, $f:A\to\mathbb{R}$ is differentiable, and $\left\lVert \nabla f(\bm{x})\right\rVert_2\le L$ throughout $A$, then $$|f(\bm{x})-f(\bm{y})|\le L\left\lVert \bm{x}-\bm{y}\right\rVert_2.$$ This is one common route from derivative bounds to continuity estimates. ## Almost everywhere, essential supremum, and essential infimum Lebesgue spaces identify functions that differ only on sets of measure zero. This motivates the **essential supremum** and **essential infimum**. A property holds **almost everywhere** (a.e.) on $\Omega$ if it fails only on a set of measure zero. In $\mathbb{R}^d$, individual points and finite or countable sets have measure zero. A curve in a three-dimensional domain also has three-dimensional measure zero. For a measurable function $f:\Omega\to\mathbb{R}$, the essential supremum is $$\operatorname*{ess\,sup}_{x\in\Omega} f(x) = \inf\{M\in\mathbb{R}:f(x)\le M\text{ for almost every }x\in\Omega\}.$$ The essential infimum is $$\operatorname*{ess\,inf}_{x\in\Omega} f(x) = \sup\{m\in\mathbb{R}:m\le f(x)\text{ for almost every }x\in\Omega\}.$$ The adjective *essential* means that exceptional values on measure-zero sets are ignored. ::: {#exm-appendix-real-essential-supremum} ## Supremum and essential supremum can differ Define $f:[0,1]\to\mathbb{R}$ by $$f(x)= \begin{cases} 100, & x=1/2,\\ 1, & x\ne 1/2. \end{cases}$$ Then $$\sup_{[0,1]}f=100, \qquad \operatorname*{ess\,sup}_{[0,1]}f=1.$$ The single exceptional point has measure zero. ::: The $L^\infty$ norm is defined through the essential supremum: $$\left\lVert f\right\rVert_{L^\infty(\Omega)} = \operatorname*{ess\,sup}_{x\in\Omega}|f(x)|.$$ This makes the norm insensitive to changes on measure-zero sets, consistent with the way $L^p$ spaces are defined. For a coefficient field $k(\bm{x})$ in an elliptic PDE, assumptions such as $$0<k_{\min} \le k(\bm{x}) \le k_{\max}<\infty \qquad\text{a.e. in }\Omega$$ are more naturally expressed as $$0<\operatorname*{ess\,inf}_{\Omega}k, \qquad \operatorname*{ess\,sup}_{\Omega}k<\infty.$$ These bounds later enter coercivity and continuity estimates. ## Sequences of functions: pointwise and uniform convergence Let $f_n:A\to\mathbb{R}$ be a sequence of functions. The sequence converges **pointwise** to $f$ if, for every fixed $x\in A$, $$f_n(x)\to f(x).$$ The index $N$ required to make $|f_n(x)-f(x)|<\varepsilon$ is allowed to depend on $x$. The sequence converges **uniformly** to $f$ if $$\sup_{x\in A}|f_n(x)-f(x)|\to0.$$ Equivalently, for every $\varepsilon>0$ there exists one $N$ that works for all $x\in A$. Uniform convergence is stronger than pointwise convergence. A uniform limit of continuous functions is continuous. A pointwise limit of continuous functions need not be continuous. ::: {#exm-appendix-real-pointwise-not-uniform} ## Pointwise but not uniform convergence On $[0,1]$, let $$f_n(x)=x^n.$$ Then $$f_n(x)\to \begin{cases} 0, & 0\le x<1,\\ 1, & x=1. \end{cases}$$ The limit is discontinuous, so the convergence cannot be uniform. ::: In finite element analysis, one often studies convergence in a norm such as $L^2$ or $H^1$, which is different from both pointwise and uniform convergence. The important habit is to always ask: *in what sense is the approximation converging?* ## A compact list of implications worth remembering For subsets of $\mathbb{R}^d$ and functions defined on them: $$\boxed{\text{compact in }\mathbb{R}^d\iff\text{closed and bounded}}$$ $$\boxed{\text{continuous on compact}\Longrightarrow\text{bounded and attains max/min}}$$ $$\boxed{\text{continuous on compact}\Longrightarrow\text{uniformly continuous}}$$ $$\boxed{\text{Lipschitz}\Longrightarrow\text{uniformly continuous}\Longrightarrow\text{continuous}}$$ $$\boxed{\max A\text{ exists}\Longrightarrow\max A=\sup A,\quad \min A\text{ exists}\Longrightarrow\min A=\inf A}$$ No reverse implication in these statements should be assumed without additional hypotheses. ## Practice problems ::: {#exr-appendix-real-sets-compactness} ## A. Sets and compactness 1. Classify each set as open, closed, both, or neither in $\mathbb{R}$: $(0,2)$, $[0,2]$, $[0,2)$, $\mathbb{R}$, $\varnothing$. 2. For $A=(0,1]\subset\mathbb{R}$, determine $A^\circ$, $\overline A$, and $\partial A$. 3. Is the set $\{1/n:n\in\mathbb N,\ n\ge1\}$ closed in $\mathbb{R}$? Is it compact? Explain. 4. Determine whether each set is compact in $\mathbb{R}^2$: the open unit disk, the closed unit disk, the line segment $\{(x,0):0\le x\le1\}$, and the whole $x$-axis. ::: ::: {#exr-appendix-real-sequences} ## B. Sequences and limits 1. Determine the limit of $x_n=(1/n,\,1-1/n)$ in $\mathbb{R}^2$. 2. Show that $x_n=(-1)^n+1/n$ does not converge, identify two convergent subsequences, and state the limit of each subsequence. 3. Determine whether $x_n=\log(n+1)-\log n$ is Cauchy. 4. Give an example of a bounded sequence that does not converge but has a convergent subsequence. ::: ::: {#exr-appendix-real-bounds} ## C. Supremum, infimum, maximum, minimum When a requested maximum or minimum does not exist, state why it does not exist. 1. For each set, determine $\sup$, $\inf$, $\max$, and $\min$ when they exist: $$(2,5),\qquad [2,5),\qquad \{1-1/n:n\ge1\},\qquad \{(-1)^n:n\ge1\}.$$ 2. Let $f(x)=x(1-x)$ on $(0,1)$ and on $[0,1]$. Determine the supremum, infimum, maximum, and minimum in each case. 3. Let $f(x)=1/x$ on $(0,1)$. Is $f$ bounded above? Does it have a minimum? ::: ::: {#exr-appendix-real-continuity} ## D. Continuity and extrema 1. Explain why $f(x)=x^2+1$ has a maximum and minimum on $[-2,3]$. 2. Does $f(x)=x^2+1$ have a maximum on $\mathbb{R}$? Does it have a minimum? 3. Give an example of a continuous function on a bounded open set that does not attain its supremum. 4. Show that $f(x)=x^2$ is Lipschitz on $[-1,1]$. Find one admissible Lipschitz constant. 5. Is $f(x)=\sqrt{x}$ Lipschitz on $[0,1]$? Is it uniformly continuous there? ::: ::: {#exr-appendix-real-essential-bounds} ## E. Essential bounds 1. Define $f:[0,1]\to\mathbb{R}$ by $f(0)=7$ and $f(x)=2$ for $x>0$. Find $\sup f$, $\inf f$, $\operatorname*{ess\,sup}f$, and $\operatorname*{ess\,inf}f$. 2. Define $f(x)=1$ if $x$ is rational and $f(x)=0$ if $x$ is irrational on $[0,1]$. Determine $\sup f$, $\inf f$, $\operatorname*{ess\,sup}f$, and $\operatorname*{ess\,inf}f$. 3. Suppose $k(x)=2+x$ for almost every $x\in(0,1)$ but $k(1/2)=1000$. Find $\operatorname*{ess\,inf}k$ and $\operatorname*{ess\,sup}k$. ::: ::: {#exr-appendix-real-function-convergence} ## F. Pointwise and uniform convergence 1. For $f_n(x)=x/n$ on $[0,1]$, determine the pointwise limit and show that the convergence is uniform. 2. For $f_n(x)=x^n$ on $[0,1]$, compute $\sup_{x\in[0,1]}|f_n(x)-f(x)|$ for the pointwise limit $f$ and explain why convergence is not uniform. 3. Let $f_n(x)=1/(n+x)$ on $[0,\infty)$. Determine whether $f_n\to0$ uniformly. ::: ## Selected answers {#sec-appendix-real-analysis-answers .unnumbered} ### A. Sets and compactness {.unnumbered} 1. $(0,2)$ open; $[0,2]$ closed; $[0,2)$ neither; $\mathbb{R}$ and $\varnothing$ both open and closed. 2. $A^\circ=(0,1)$, $\overline A=[0,1]$, $\partial A=\{0,1\}$. 3. The set is not closed because $0$ is a limit point not contained in it; therefore it is not compact. 4. Only the closed unit disk and the line segment are compact. ### C. Supremum, infimum, maximum, minimum {.unnumbered} 1. $(2,5)$: $\sup=5$, $\inf=2$, no max/min. $[2,5)$: $\sup=5$, $\inf=\min=2$, no max. For $\{1-1/n\}$: $\sup=1$, no max, $\inf=\min=0$. For $\{(-1)^n\}$: $\sup=\max=1$, $\inf=\min=-1$. 2. On both domains, $\sup=\max=1/4$ at $x=1/2$. On $(0,1)$, $\inf=0$ but no minimum; on $[0,1]$, $\inf=\min=0$. 3. Unbounded above; minimum $1$ is not attained on $(0,1)$, so there is no minimum and $\inf=1$. ### E. Essential bounds {.unnumbered} 1. $\sup=7$, $\inf=2$, $\operatorname*{ess\,sup}=\operatorname*{ess\,inf}=2$. 2. $\sup=1$, $\inf=0$, $\operatorname*{ess\,sup}=0$, $\operatorname*{ess\,inf}=0$ because the rationals have measure zero. 3. $\operatorname*{ess\,inf}k=2$ and $\operatorname*{ess\,sup}k=3$.

A.1 Sets in \(\mathbb{R}^d\): balls, neighborhoods, interior, closure, and boundary