← Regular and Normal Spaces
What should you be acquainted with? Analysis and basics of Functional Analysis.

Compact and Locally Compact Spaces

Lindelöf Spaces and Compact Spaces

A Hausdorff space $X$ is said to be compact, if every open cover of $X$ contains a finite subcover. A Hausdorff space $X$ is said to be a Lindelöf space, if every open cover contains a countable subcover.
A Hausdorff space with countable basis is a Lindelöf space.
Let $X$ be a regular Lindelöf space. Then for every open cover $(U_\a)$, $\a\in I$, we can find a locally finite countable subcover $W_k$, such that every $W_k$ is contained in some $U_\a$. Moreover $X$ is normal.
A Hausdorff space $X$ is compact if and only if every family $A_\a$ of closed, nonempty subsets with the property that each finite subfamily has non-empty intersection (this is called the finite intersection property), has non-empty intersection.
For a Hausdorff space $X$ the following assertions are equivalent: 1. $X$ is compact. 2. Every filter on $X$ has an accumulation point. 3. For every filter on $X$ there is a finer filter, which converges. 4. Every ultra filter on $X$ converges.
Suppose $A,B$ are compact and disjoint subsets of a Hausdorff space $X$. Then there exist open, disjoint neighborhoods $U$ and $V$, such that $A\sbe U$ und $B\sbe V$. In particular: every compact space is normal.
1. Every compact subset of a Hausdorff space is closed. 2. Every closed subset $A$ of a compact space is compact.
Let $X,Y$ be Hausdorff and $f:X\rar Y$ continuous. If $X$ is compact, then $f(X)$ is compact.
Suppose $X$ is compact, $Y$ Hausdorff and $f:X\rar Y$ any mapping. 1. If $f$ is continuous and bijectiv, then $f$ is a homeomorphism. 2. $f$ is continuous if and only if its graph $\G(f)$ is compact.
Suppose $X$ is compact, $Y$ Hausdorff and $f:X\rar Y$ a continuous surjection. Then $Y$ is homeomorphic to $X/R$ where $x_1Rx_2$ iff $f(x_1)=f(x_2)$.
If $f:X\rar\cl\R$ l.s.c on a compact space $X$, then $f$ attains its infimum.
Obviously, the disjoint union $\coprod X_\a$, $\a\in I$, is compact if and only if $I$ is finite.
Let $X_\a$, $\a\in I$, be a family of compact spaces. Than $\prod X_\a$ is compact.
Let $X$ be a compact space and $Z(x)$ the connected component of a point $x\in X$. Then $$ Z(x)=\bigcap\{B\sbe X:\,x\in B, B^\circ=\cl B\} $$ 2. If $A\sbe X$ is closed and $A\cap Z(x)=\emptyset$, then there is a closed and open (clopen) subset $B$ von $X$, such that $Z(x)\sbe B$ and $A\cap B=\emptyset$.
A compact connected manifold of dimension one is homeomorphic to $S^1$.

Applications in Functional Analysis
Let $X$ be a compact space and $A$ a subalgebra of the space $C(X)$ of continuous, real valued functions on $X$ with norm $\norm f\colon=\sup\{|f(x)|:x\in X\}$. If $A$ separates points of $X$ and if for all $x\in X$ there is some $f\in A$ such that $f(x)\neq0$, then $A$ is dense in $C(X)$.
Let $X$ be a Banach space with dual $X^*$; the weak and the weak-* topology on $X$ and $X^*$ respectively is the initial topology on $X$ and $X^*$ respectively with respect to the mappings $x\mapsto x^*(x)$, $x^*\in X^*$ and $x^*\mapsto x^*(x)$, $x\in X$. These topologies will be denoted by $\s(X,X^*)$ and $\s(X^*,X)$. In order to emphasize the reciprocity of $X$ and $X^*$ the notation $\la x,x^*\ra\colon=x^*(x)$ is commonly used.
For every Banach space $X$ the space $(\cl B_{X^*},\s(X^*,X))$ is compact. 2. If $X$ is separable, the $(\cl B_{X^*},\s(X^*,X))$ is metrizable. 3. $B_X$ is a dense subset of $(\cl B_{X^{**}},\s(X^{**},X^*))$.
A Banach space $X$ is reflexive, if and only if the closed unit ball $\cl B_{X}$ of $X$ is a compact subset of $(X,\s(X,X^*))$.
The closed unit ball $\cl B_{X}$ of $X$ equipped with the weak topology is homeomorphic to a subset of $K^{B_X}$ ($K$ denotes the closed unit ball of $\R$ or $\C$), but this subset need not be closed.
For a Banach space $X$ define $\d:\R^+\rar\R_0^+$ by $$ \d(t)\colon=\inf\{1-\norm{x+y}/2:\, \Vert x\Vert,\norm y\leq1,\norm{x-y}\geq t\}~. $$ This is called the modulus of convexity of $X$. If $\d(t)>0$, then we say that $X$ is uniformly convex. Every uniformly convex space is reflexive.
For $1 < p < \infty$ the space $L_p(\mu)$ is uniformly convex and thus reflexive.
Suppose $f:X\rar(-\infty,\infty]$ is a convex l.s.c. function on a Banach space $X$, then $f$ is also l.s.c. with respect to the weak topology, because for all $t\in\R$ the set $[f\leq t]$ is convex and norm-closed; by the Hahn-Banach Theorem this set is also closed with respect to the weak topology. Now assume, that $X$ is reflexive and $[f\leq t]$ is non-empty and bounded for some $t\in\R$. Then there is a point $x_0\in[f\leq t]$, such that $f(x_0)=\inf\{f(x):x\in X\}$. Moreover, if $f$ is stricly convex on $[f\leq t]$ then $x_0$ is unique.

Compact Metric Spaces
A Hausdorff space $X$ is said to be sequentially compact, if every sequence in $X$ contains a covergent subsequence.
Let $X$ a metric space or a Hausdorff space with countable basis. Then $X$ is compact if and only if $X$ is sequentially compact.
A subset $A$ of a Hausdorff space $X$ is said to be relatively compact, if its closure is compact. A metric space $(X,d)$ is said to be pre-compact, if for every $\e>0$ we can find a finite subset $A$ of $X$, such that for all $x\in X$: $d(x,A)<\e$.
A metric space $X$ is compact, if and only if $X$ is complete and pre-compact.
Let $(X,d)$ be a complete metric space and $A\sbe X$. $A$ is relatively compact if and only if for any $r>0$ we can find a relatively compact subset $K_r$ of $X$, such that $A\sbe[d_{K_r}< r]$. In particular, if $X=E$ is a complete metrizable vector space and $B_r\colon=B_r(0)$, then $A\sbe E$ is relatively compact, if and only if for any $r>0$ there is a relatively compact subset $K_r$ of $E$, such that $A\sbe K_r+B_r$.
A minor but useful generalization is the following
Let $K_n$ be compact subsets of a complete metric space $X$ and suppose that $J_n:X\rar K_n$ and $I_n:K_n\rar X$ are uniformly continuous. If $$ \lim_{n\to\infty}\sup\left\{d(I_n(J_n(x)),x):x\in X\right\}=0~. $$ then $X$ is compact.
Let $(X,d)$ be a compact metric space and $U_1,\ldots,U_n$ an open cover of $X$. Then there exists some $\d>0$, such that for any subset $A$ of $X$ satisfying $d(A)\leq\d$, there exists some index $k$ such that $A\sbe U_k$. The largest numer $\d$ with this property is called the Lebesgue number of the cover.

Compactness Criteria
A regular space $X$ is compact if and only if for every Hausdorff space $Y$ the projection $\Prn_Y:X\times Y\rar Y$ is closed.
Let $X$ be compact and $R\sbe X\times X$ a closed equivalence relation. Then $X/R$ is compact. If moreover $X$ is metrizable, then so is $X/R$.
Suppose that $X$ is Hausdorff and $Y$ compact. $f:X\rar Y$ is continuous if and only if $\G(f)$ is closed.
If $X$ is compact and $F:X\times Y\rar\R$ is a continuous function, then $$ f(y)\colon=\sup\{F(x,y):\,x\in X\} $$ is continuous: $f:Y\rar\R$ is l.s.c. and thus we only have to show that $f$ is u.s.c., i.e. for all $t\in\R$ the set $[f\geq t]$ is closed, but since $X$ is compact this set coincides with $\Prn_Y([F\geq t])$. This result essentially shows the necessity parts of the following two criteria.
Suppose $X$ is compact. A subset $H$ of $C(X)$ with norm $\norm f\colon=\sup\{|f(x)|:x\in X\}$ is relatively compact if and only if it is bounded and for every $x\in X$ the mapping $$ y\mapsto\sup\{|f(x)-f(y)|:f\in H\} $$ is continuous - we say $H$ is equicontinuous at $x$.
Let $1\leq p<\infty$, $\O$ an open subset of $\R^n$ and $K_m$ an increasing sequence of compact subsets of $\O$ such that $\bigcup K_m=\O$.
A subset $H$ of $L_p(\O)$ is relatively compact if and only if it is bounded and $$ \lim_n\sup_{f\in H}\int_{K_n^c}|f(x)|^p\,dx=0, \quad \lim_{y\to0}\sup_{f\in H}\int_{K_n}|f(x-y)-f(x)|^p\,dx=0 $$
This theorem generalizes to Riemannian manifolds: Suppose $M$ is a compact Riemannian manifold and $G$ denotes the flow of the geodesic vector field ${\cal X}$ on the tangent space $TM$, i.e. for each tangent vector $X_m$ at $m$ the mapping $\g:t\mapsto G(t,X_m)$ is the geodesic through $m$ satisfying $\g^\prime(0)=X_m$. A bounded subset $H\sbe L_p(M)$ is relativly compact if and only if for all vector fields $X$ on $M$: $$ \lim_{t\to0}\sup_{f\in H} \int|f\circ\pi(G(t,X_m))-f(m)|^p\,v(dm)=0~. $$ Here $v$ denotes the Riemannian volume on $M$ and $\pi:TM\rar M$ the cannonical projection that sends each tangent vector $X_m$ at $m$ to $m$. In case $H$ is a subset of $C^\infty(M)$ it's obvious to interpret this condition in terms of derivatives. For simplicity we stick to the Euclidean case, i.e. $M=\R^n$. By Jensen's inequality we have: $|f(x+y)-f(x)|^p\leq\int_0^1|df(x+ty)y|^p\,dt$; furthermore by Fubini and translation invariance of the Lebesgue measure: $\int|f(x+y)-f(x)|^p\,dx\leq\norm y^p\int\norm{\nabla f}^p\,d\l$. By theorem $H$ is relatively compact in $L_p(\R^n)$ if $$ \sup_{f\in H}\int|f|^p+\norm{\nabla f}^p\,d\l<\infty \quad\mbox{and}\quad \lim_n\sup_{f\in H}\int_{K_n^c}|f|^p\,d\l=0~. $$ Suppose $\mu$ is a Radon measure on $\R^n$ with density $\r=e^{-U}$ with respect to Lebesgue measure. Then the mapping $J: f\mapsto fe^{-U/p}$ is an isometry from $L_p(\mu)$ onto $L_p(\R^n)$ and thus a subset $H$ von $L_p(\mu)$ is relatively compact, if and only if $J(M)$ is relatively compact in $L_p(\R^n)$.
Let $\D\colon=-\sum\pa_j^2$ be the Laplacian on $\R^n$. If $\lim_{x\to\infty}(2\D U(x)+\norm{\nabla U(x)}^2)=+\infty$, then for any $C<\infty$ the set $$ B_C\colon=\Big\{f\in C_c^\infty(\R^n): \norm f_1^2\colon=\int f^2+\norm{\nabla f}^2\,d\mu\leq C\Big\}~. $$ is relatively compact in $L_2(\mu)$. The condition on $U$ is equivalent to: $-\D\sqrt\r/\sqrt\r$ converges to $+\infty$ for $x\to\infty$.
If $X$ is a complete metric space, then so is ${\cal H}(X)$.
If $X$ is a compact metric space, then so is ${\cal H}(X)$.
If $p\in(1,\infty)$, then a subset of $L_p(\O,\F,\mu)$ is weakly relatively compact, if and only if it is bounded. The situation is of course different in the case $p=1$.
A bounded subset $H$ of $L_1(\O,\F,\mu)$ is weakly relatively compact if and only if the following conditions are satisfied: 1. for any $\e>0$ there exists a set $A\in\F$, such that $\mu(A)<\infty$ and $$ \sup_{f\in H}\int_{A^c}|f|\,d\mu\leq\e~. $$ 2. $H$ is uniformly integrable, i.e. $$ \lim_{t\to\infty}\sup_{f\in H}\int_{[|f|>t]}|f|\,d\mu=0~. $$

Locally Compact Spaces
A Hausdorff space $X$ is said to be locally compact, if each point $x\in X$ has a neighborhood basis consisting of compact sets.
Suppose $X$ locally compact, then for every compact subset $K$ there exists an open superset $U$ and a continuous function $f:X\rar [0,1]$ such that $f|K=1$ and $f|U^c=0$. If $X$ is a smooth manifold $f$ can be chosen to be smooth.
Let $X$ be locally compact, $K$ a compact subset of $X$ and $V_j$, $1\leq j\leq n$ open subsets such that $K\sbe\bigcup V_j$. Then there exist continuous functions $\psi_j:X\rar[0,1]$ satisfying $\supp(\psi_j)\sbe V_j$ and $\sum\psi_j|K=1$. If $X$ is a smooth manifold the functions $\psi_j$ can be chosen to be smooth.
The support $\supp(f)$ of a function $f:X\rar\R$ is the smallest closed subset containing $[f\neq0]$.
Let $X$ be locally compact, $Y$ Hausdorff and let $f:X\rar Y$ be continuous open and onto. Then for every compact subset $K$ of $Y$ there exists a compact subset $C$ of $X$, such that $K=f(C)$.
Suppose $X$ is locally compact, metrizable and connected. Then there is a sequence $C_n$ of compact subsets, such that for every compact subset $C$ there exists some $n\in\N$ such that $C\sbe C_n$. In particular, $X$ is separable.
Thus for a topological manifold the space $C(M)$ with the topology of uniform convergence on compact sets is a complete metric space. Moreover in case $M$ is a smooth manifold the space $C^\infty(M)$ of all smooth (real- or complex valued) functions with the topology of uniform convergence of all derivatives on compact sets is a complete metric space. By the Stone-Weierstraß Theorem a subalgebra ${\cal A}$ von $C(M)$ is dense if it seperates points and if for all $x\in M$ there is some $f\in$ such that $f(x)\neq0$. For $C^\infty(M)$ there is the following
Let $M$ be a smooth manifold and ${\cal A}$ a subalgebra of the space $C^\infty(M)$ of continuous, real valued functions on $M$. Suppose 1. ${\cal A}$ separates points of $M$, 2. for every $x\in M$ there is some $f\in{\cal A}$ such that $f(x)\neq0$ and 3. for every $x\in M$ and every $u\in T_xM\sm\{0\}$ there is some $f\in{\cal A}$ such that $df(x)u\neq0$. Then ${\cal A}$ is dense in $C^\infty(M)$.

Vitali Covering Theorem

In the sequel let $\mu$ be a Radon measure on a locally compact, separable, metric space $(X,d)$. We will also assume that there is some constant $C<\infty$ such that for all $x\in X$ and all $r>0$: $$ \mu(B_r^\prime(x))\geq v(r) \quad\mbox{and}\quad \mu(B_{3r}^\prime(x))\leq C\mu(B_r^\prime(x)), $$ where $v:\R^+\rar\R^+$ denotes some increasing function and $B_r^\prime(x)$ the closed ball $\{y:d(x,y)\leq r\}$.
Suppose $A$ is a measurable subset of $X$ and $\B$ a family of bounded balls covering $A$. Then we can find a pairwise disjoint sequence $D_n$ in $\B$, such that $C\sum\mu(D_n)\geq\mu(A)$.
Let $f$ be locally integrable on $X$, $r>0$ and put $$ f_r(x)\colon=\frac1{\mu(B_r^\prime(x))} \int_{B_r^\prime(x)}f\,d\mu \quad\mbox{und}\quad Mf\colon=\sup_{r > 0}|f|_r $$ $Mf$ is called the maximal function of $f$ and the following assertions hold:
1. If $f\in L_1(\mu)$, then for all $t>0$: $\mu(Mf>t)\leq Ct^{-1}\int|f|\,d\mu$.
2. If $f\in L_p(\mu)$, $1< p\leq\infty$, then: $\norm{Mf}_p\leq C_p\norm f_p$.

Compactification

One Point Compactification

Let $(X,{\cal T})$ be locally compact and let $\o$ be any point not in $X$. On the set $X^*\colon=X\cup\{\o\}$ we define a neighborhood basis ${\cal U}(\o)$ for $\o$: $$ U\sbe X^*\mbox{ open }:\Lrar\ \o\in U, U^c(\sbe X) \mbox{ compact} $$ Now denote by ${\cal T}^*$ the union of the collections ${\cal T}$ and ${\cal U}(\o)$. Since $X$ is locally compact, the space $X^*$ is Hausdorff and $(X^*,{\cal T}^*)$ is compact; moreover, the canonical injection $X\rar X^*$ is an embedding. $X^*$ is called one-point-compactification or Alexandrov compactification of $X$. The space $C_0(X)$ is the vector space of all continuous functions $f:X\rar\C$ such that for every $\e>0$ there exists some $U\in{\cal U}(\o)$ such that $|f||U<\e$. $C_0(X)$ equipped with the norm $\norm f\colon=\sup\{|f(x)|:x\in X\}$ is a Banach space and the canonical injection $C_0(X)\rar C(X^*)$ is an isometric embedding; moreover $C(X^*)=C_0(X)\oplus[1]$.
1. A subset $H\sbe C_0(X)$ is relatively compact, if and only if $H$ is bounded, equicontinuous and for each $\e>0$ there is a compact subset $K$ of $X$, such that $\sup\{|f(x)|:f\in H, x\notin K\}<\e$.
2. The Stone-Weierstraß Theorem: A point separating subalgebra ${\cal A}\sbe C_0(X)$ with the property that for every $x\in X$ there exists some $f\in{\cal A}$, such that $f(x)\neq0$, is dense in $C_0(X)$.

Standard Compactification of $\R$

The extended real numbers $\cl{\R}\colon=\R\cup\{\pm\infty\}$ are homeomorphic to the compact $[-1,1]$; a homeomorphism $\vp:\cl{\R}\rar[-1,1]$ is given by e.g. $$ \vp(x)=\arctan(x), \vp(-\infty)=-1, \vp(+\infty)=+1~. $$ Let $\F$ be a basis of a filter on the topological space $X$ and suppose $f:X\rar\cl\R$ is any function. The set $A$ of accumulation points of $f(\F)$ is then given by $A=\bigcap\{\cl{f(F)}:F\in\F\}$. By compactness of $\cl{\R}$ this set is not empty; we define $$ \limsup f(\F)\colon=\sup A \quad\mbox{und}\quad \liminf f(\F)\colon=\inf A~. $$ For a sequence (or a net) $x_n$ in $X$, we put $F_n\colon=\{x_m:m\geq n\}$ and $\F=\{F_n:n\in\N\}$. Then we have $$ \limsup f(\F)=\inf_n\sup_{m\geq n}f(x_m) \quad\mbox{und}\quad \liminf f(\F)=\sup_n\inf_{m\geq n}f(x_m)~. $$

Stone Čech Compactification

Suppose $X$ is completely regular. Let $C(X,[0,1])$ be the set of continuous functions $f:X\rar[0,1]$. The mapping $$ \Phi_X:X\rar[0,1]^{C(X,[0,1])}=\colon S_X \qquad x\mapsto(f(x))_{f\in C(X,[0,1])} $$ is a homeomorphism from $X$ onto its image $\Phi_X(X)$.
The Stone Čech compactification $\b X$ of $X$ is the closure of $\Phi_X(X)$ in $S_X$. A compactification of $X$ is an embedding $j:X\rar cX$ in a compact space $cX$, such that $j(X)$ is dense in $cX$.
If $X$ is completely regular, $K$ compact and $f:X\rar K$ continuous, then there is exactly one continuous mapping $F:\b X\rar K$, such that $F\circ\Phi_X=f$. If $j:X\rar cX$ is a compactification of $X$ such that any continuous mapping $f:X\rar K$ into any compact space $K$ admits a continuation $F:cX\rar K$, $F\circ j=f$, then $cX$ is homeomorphic to $\b X$.
Let $A$ be a dense subset of a Hausdorff space $X$ and $f:X\rar Y$ continuous. If $f:A\rar f(A)$ is a homoemorphism, then: $f(A^c)\cap f(A)=\emptyset$.
Suppose $j_1:X\rar K_1$ and $j_2:X\rar K_2$ are compactifications of $X$. Then for every continuous mapping $f:K_1\rar K_2$ satisfying $f\circ j_1=j_2$ we have: $$ f(j_1(X))=j_2(X) \quad\mbox{and}\quad f(K_1\sm j_1(X))=K_2\sm j_2(X)~. $$
Let $X$ be a separable, metric space. Then there exists a compact metric space $cX$ containing $X$ as a dense subset.
For a completely regular space $X$ the Banach space $C_b(X)$ is isometrically isomorphic to $C(\b X)$: we just assign every continuous bounded function $f\in C_b(X)$ its continuation $F\in C(\b X)$.
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Last modified: Mon Sep 18 18:45:34 CEST 2023