Topology/Compactness

The notion of Compactness appears in a wide variety of contexts. In particular, compactness is a "tameness property" that tells you that the objects you are dealing with are in some sense well-behaved.

Let ${\displaystyle X}$ be a topological space and let ${\displaystyle S\subseteq X}$

A collection ${\displaystyle {\mathcal {G}}}$ of open sets ${\displaystyle \{G_{\alpha }\}}$ is said to be an Open Cover of ${\displaystyle S}$ if ${\displaystyle S\subseteq \displaystyle \bigcup _{\mathcal {G}}G_{\alpha }}$

${\displaystyle S}$ is said to be Compact if and only if every open cover of ${\displaystyle S}$ has a finite subcover. More formally, ${\displaystyle S}$ is compact iff for every open cover ${\displaystyle {\mathcal {G}}}$ of ${\displaystyle S}$, there exists a finite subset ${\displaystyle \{G_{\alpha _{1}},G_{\alpha _{2}},\ldots ,G_{\alpha _{n}}\}}$ of ${\displaystyle {\mathcal {G}}}$ that is also an open cover of ${\displaystyle S}$.

If the set ${\displaystyle X}$ itself is compact, we say that ${\displaystyle X}$ is a Compact Topological Space.

Compactness of topological spaces can also be expressed by one of the following equivalent characterisations:

• Every filter on ${\displaystyle X}$ containing a filter basis of closed sets has a nonempty intersection.
• Every ultrafilter on ${\displaystyle X}$ converges.

• Every closed subset of a compact set is compact
  Proof:
  Let ${\displaystyle S\subseteq X}$ be a compact set, and let ${\displaystyle K}$ be a closed subset of ${\displaystyle S}$. Consider any open cover ${\displaystyle {\mathcal {G}}}$ of ${\displaystyle K}$. Observe that ${\displaystyle X\setminus K}$ being open, the collection of open sets ${\displaystyle {\mathcal {G}}\cup \{X\setminus K\}}$is an open cover of ${\displaystyle S}$. As ${\displaystyle S}$ is compact, this open cover has a finite subcover ${\displaystyle {\mathcal {G}}'}$.
  Now, consider the collection ${\displaystyle {\mathcal {G}}'\setminus \left(X\setminus K\right)}$. This collection is obviously finite and is also a subcover of ${\displaystyle {\mathcal {G}}}$. Hence, it is a finite subcover of ${\displaystyle K}$
  

• Every compact subset of a Hausdorff space is closed.
  Proof:
  Let ${\displaystyle K}$ be compact. If the complement ${\displaystyle K^{c}}$ is empty, then ${\displaystyle K}$ is the same as the space; thus closed. Suppose not; that is, there is a point ${\displaystyle x\in K^{c}}$. Then for each ${\displaystyle y\in K}$, by the Hausdorff separation axiom we can find ${\displaystyle U_{y}}$ and ${\displaystyle V_{y}}$ disjoint, open and such that ${\displaystyle x\in U_{y}}$ and ${\displaystyle y\in V_{y}}$. Since ${\displaystyle K}$ is compact and the collection ${\displaystyle \{V_{y};y\in K\}}$ covers ${\displaystyle K}$, we can find a finite number of points ${\displaystyle y_{1},y_{2},...y_{n}}$ in ${\displaystyle K}$ such that:
  ${\displaystyle K\subseteq V_{y_{1}}\cup V_{y_{2}}\cup \ldots \cup V_{y_{n}}}$
  It then follows that:
  ${\displaystyle x\in U_{y_{1}}\cap U_{y_{2}}\cap \ldots \cap U_{y_{n}}=U_{x}}$. Hence, every ${\displaystyle x\in X}$ has an open neighbourhood ${\displaystyle U_{x}}$.
  As ${\displaystyle K^{c}}$ can be represented as the union of open sets ${\displaystyle U_{x}}$, ${\displaystyle K^{c}}$ is open and ${\displaystyle K}$ is closed.

• Every compact set in a metric space is bounded.
  Proof:
  Let ${\displaystyle X}$ be a metric space and let ${\displaystyle K\subset X}$ be compact.
  Consider the collection of open balls ${\displaystyle {\mathcal {G}}=\{B_{r}(p)|r\in \mathbb {R} ^{+}\}}$ for some (fixed) ${\displaystyle p\in X}$. We see that ${\displaystyle {\mathcal {G}}}$ is an open cover of ${\displaystyle K}$. As ${\displaystyle K}$ is compact, it has a finite subcover, say ${\displaystyle \{B_{r_{1}}(p),B_{r_{2}}(p),\ldots ,B_{r_{n}}(p)\}}$. Let ${\displaystyle r_{0}=\sup\{r_{1},r_{2},\ldots ,r_{n}\}}$. We see that ${\displaystyle K\subset B_{r_{0}}(p)}$, and hence, ${\displaystyle K}$ is bounded.

• Heine-Borel Theorem: For any interval ${\displaystyle [a,b]}$, and for any open cover ${\displaystyle {\mathcal {G}}}$ of that interval, there exists a finite subcover of ${\displaystyle {\mathcal {G}}}$.
  Proof:
  Let ${\displaystyle S}$ be the set of all ${\displaystyle a\leq x\leq b}$ such that ${\displaystyle [a,x]}$ has a finite subcover of ${\displaystyle {\mathcal {G}}}$. ${\displaystyle S}$ is non-empty because ${\displaystyle a}$ is within the set. Define ${\displaystyle c=\sup S}$.
  Assume if possible, ${\displaystyle c<b}$. Then there is a finite cover of sets within ${\displaystyle {\mathcal {G}}}$ for ${\displaystyle [a,c]}$. ${\displaystyle c}$ is within a set ${\displaystyle A}$ within the cover ${\displaystyle {\mathcal {G}}}$. Thus, there exists a ${\displaystyle \varepsilon >0}$ such that ${\displaystyle (c-\varepsilon ,c+\varepsilon )\subseteq A}$. Then ${\displaystyle c+{\tfrac {\varepsilon }{2}}}$ is also within ${\displaystyle S}$, contradicting the definition of ${\displaystyle c}$. Thus, ${\displaystyle c\geq b}$. Therefore, ${\displaystyle {\mathcal {G}}}$ has a finite subcover.

Sources differ as to what exactly should be called the 'Heine-Borel Theorem'. It seems that Emile Borel proved the most relevant result, dealing with compact subsets of a Euclidean Space. However, we provide the simpler case, for reals.

• Let ${\displaystyle X,Y}$ be topological spaces. If ${\displaystyle f:X\to Y}$ is continuous, and ${\displaystyle A\subset X}$ is compact, then the image of ${\displaystyle A}$, ${\displaystyle f(A)}$, is compact.
  Proof:
  Let ${\displaystyle {\mathcal {G}}}$ be any open cover of ${\displaystyle f(A)}$. Consider the inverses ${\displaystyle \{f^{-1}(B)\}}$ where ${\displaystyle B\in {\mathcal {G}}}$. These inverses are open because ${\displaystyle f}$ is continuous. This covers ${\displaystyle A}$, and thus there is a finite subcover of ${\displaystyle A}$, ${\displaystyle \{B_{i}\}}$. Then the images ${\displaystyle \{f(B_{i})\}}$ is a finite subcover of ${\displaystyle f(A)}$.

• If a set is compact and Hausdorff, then it is normal.
  Proof:
  Let ${\displaystyle X}$ be compact and Hausdorff. Consider two closed subsets ${\displaystyle A}$ and ${\displaystyle B}$ which are themselves compact by theorem 1 above. For every ${\displaystyle a\in A}$ and ${\displaystyle b\in B}$, there exist two disjoint sets ${\displaystyle O_{a,b,1}}$ and ${\displaystyle O_{a,b,2}}$ such that ${\displaystyle a\in O_{a,b,1}}$ and ${\displaystyle b\in O_{a,b,2}}$. The union of all such ${\displaystyle O_{a,b,2}}$ for a fixed ${\displaystyle a}$ is a cover for ${\displaystyle B}$, and thus it has a finite subcover, say, ${\displaystyle {\mathcal {O}}_{a,2}}$ and let ${\displaystyle O'_{a,2}}$ be the union of its members.
  Let ${\displaystyle B_{a}=\{b|O_{a,b,2}\in {\mathcal {O}}_{a,2}\}}$, and let ${\displaystyle O_{a,1}=\displaystyle \bigcap _{b\in B_{a}}O_{a,b,1}}$. Observe that ${\displaystyle B_{a}}$ being finite, ${\displaystyle O_{a,1}}$ is open. The union ${\displaystyle \displaystyle \bigcup _{a\in A}O_{a,1}}$ covers ${\displaystyle A}$, and therefore it has a finite subcover ${\displaystyle {\mathcal {O}}_{a,1}}$. Let ${\displaystyle U}$ be the union of all members of this subcover.
  Let ${\displaystyle A'}$ denote the set of all elements ${\displaystyle a\in A}$ such that ${\displaystyle O_{a,1}\in {\mathcal {O}}_{a,1}}$. Take the intersection ${\displaystyle V=\displaystyle \bigcap _{a\in A'}O'_{a,2}}$, which is open.
  Then ${\displaystyle U}$ is an open superset of ${\displaystyle A}$, ${\displaystyle V}$ is an open superset of ${\displaystyle B}$, and they are disjoint. Thus, ${\displaystyle X}$ is normal.

• In a compact metric space X, a function from X to Y is uniformly continuous if and only if it is continuous.
  Proof:
  

• If two topological spaces are compact, then their product space is also compact.
  Proof:
  Let X₁ and X₂ be two compact spaces. Let S be a cover of X₁×X₂. Let x be an element of X₁. Consider the sets A_x,y within S that contain (x,y) for each y in X₂. ${\displaystyle \pi _{2}:(A_{(}x,y))}$ forms a cover for X₂, with a finite subcover {A_x,yi}. Let B_x be the intersection of ${\displaystyle \pi _{1}:(A_{y})}$ within {A_yi}, which is open. Thus, {B_x} forms an open cover, which has a finite subcover, {B_xi}. The corresponding sets {A_xi,yi} is finite, and forms an open subcover of the set.

• All closed and bounded sets in the Euclidean Space are compact.
  Proof:
  Let S by any bounded closed set in ${\displaystyle R^{n}}$. Then since S is bounded, it is contained in some "box" of the products of closed intervals of R. Since those closed intervals are compact, their product is also compact. Therefore, S is a closed set in a compact set, and is therefore also compact.

The more general result on the compactness of product spaces is called Tychonoff's Theorem. Unlike the compactness of the product of two spaces, however, Tychonoff's Theorem requires Zorn's Lemma. (In fact, it is equivalent to Axiom of Choice.)

Theorem: Let ${\displaystyle X=\prod _{i\in I}X_{i}}$, and let each ${\displaystyle X_{i}}$ be compact. Then the X is also compact.

Proof: The proof is in terms of nets. Recall the following facts:

Lemma 1 - A net ${\displaystyle (x_{\alpha })_{\alpha \in {\mathcal {A}}}}$ in ${\displaystyle \prod _{i\in I}X_{i}}$ converges to ${\displaystyle x\in \prod _{i\in I}X_{i}}$ if and only if each coordinate ${\displaystyle (x_{\alpha }^{i})_{\alpha \in {\mathcal {A}}}}$ converges to ${\displaystyle x^{i}\in X_{i}}$.

Lemma 2 - A topological space ${\displaystyle X}$ is compact if and only if every net in ${\displaystyle X}$ has a convergent subnet.

Lemma 3 - Every net has a universal subnet.

Lemma 4 - A universal net ${\displaystyle (x_{\alpha })_{\alpha \in {\mathcal {A}}}}$ in a compact space ${\displaystyle X}$ is convergent.

We now prove Tychonoff's theorem.

Let ${\displaystyle (x_{\alpha })_{\alpha \in {\mathcal {A}}}}$ be a net in ${\displaystyle \prod _{i\in I}X_{i}}$.

Using Lemma 3 we can find a universal subnet ${\displaystyle (y_{\beta })_{\beta \in {\mathcal {B}}}}$ of ${\displaystyle (x_{\alpha })_{\alpha \in {\mathcal {A}}}}$.

It is easily seen that each coordinate net ${\displaystyle (y_{\beta }^{i})_{\beta \in {\mathcal {B}}}}$ is a universal net in ${\displaystyle X_{i}}$.

Using Lemma 4 we see that each coordinate net converges, because ${\displaystyle X_{i}}$ is compact.

Using Lemma 1 we see that the whole net ${\displaystyle (y_{\beta })_{\beta \in {\mathcal {B}}}}$ converges in ${\displaystyle \prod _{i\in I}X_{i}}$.

We conclude that every net in ${\displaystyle \prod _{i\in I}X_{i}}$ has a convergent subnet, so, by Lemma 2, ${\displaystyle \prod _{i\in I}X_{i}}$ must be compact. ${\displaystyle \square }$

Relative compactness is another property of interest.

Definition: A subset S of a topological space X is relative compact when the closure Cl(x) is compact.

Note that relative compactness does not carry over to topological subspaces. For example, the open interval (0,1) is relatively compact in R with the usual topology, but is not relatively compact in itself.

The idea of local compactness is based on the idea of relative compactness.

If, in a topological space X, every element has a neighborhood that is relatively compact, then X is locally compact.

It can be shown that all compact sets are locally compact, but not conversely.

Definition. Let ${\displaystyle (X,d)}$ be a metric space and let ${\displaystyle A\subset X}$ be nonempty. For ${\displaystyle x\in X}$, define the *distance from ${\displaystyle x}$ to ${\displaystyle A}$* by

${\displaystyle d(x,A)\;=\;\inf\{\,d(x,a)\mid a\in A\,\}\in [0,\infty ).}$

(If one ever needs the empty set, the standard convention is ${\displaystyle d(x,\varnothing )=+\infty }$; we will not use this case below.)

Fact (1-Lipschitz continuity of the distance-to-a-set map). Fix ${\displaystyle A\neq \varnothing }$. The function ${\displaystyle x\mapsto d(x,A)}$ is continuous and, in fact,

${\displaystyle {\bigl |}d(x,A)-d(y,A){\bigr |}\;\leq \;d(x,y)\qquad {\text{for all }}x,y\in X.}$

Proof. For any ${\displaystyle a\in A}$, the triangle inequality gives

${\displaystyle d(x,a)\;\leq \;d(x,y)+d(y,a).}$

Taking the infimum over ${\displaystyle a\in A}$ on the right yields

${\displaystyle d(x,A)\;\leq \;d(x,y)+d(y,A)\quad \Longrightarrow \quad d(x,A)-d(y,A)\;\leq \;d(x,y).}$

Interchanging ${\displaystyle x}$ and ${\displaystyle y}$ gives ${\displaystyle d(y,A)-d(x,A)\leq d(x,y)}$. Combining the two inequalities yields the desired absolute-value bound, hence continuity. ${\displaystyle \square }$

Notation. For ${\displaystyle x\in X}$ and ${\displaystyle \varepsilon >0}$, the open ${\displaystyle \varepsilon }$-ball is

${\displaystyle B(x,\varepsilon )=\{y\in X:d(x,y)<\varepsilon \}.}$

Definition (diameter). If ${\displaystyle B\subset X}$ is bounded, its diameter is

${\displaystyle \operatorname {diam} (B)=\sup\{\,d(b_{1},b_{2})\mid b_{1},b_{2}\in B\,\}\in [0,\infty ).}$

Lemma (Lebesgue number lemma). Let ${\displaystyle {\mathcal {A}}}$ be an open cover of a metric space ${\displaystyle (X,d)}$. If ${\displaystyle X}$ is compact, then there exists ${\displaystyle \delta >0}$ such that for **every** subset ${\displaystyle B\subset X}$ with ${\displaystyle \operatorname {diam} (B)<\delta }$, there exists some ${\displaystyle A\in {\mathcal {A}}}$ with ${\displaystyle B\subset A}$. Any such ${\displaystyle \delta }$ is called a *Lebesgue number* for ${\displaystyle {\mathcal {A}}}$.

Proof. We proceed in explicit steps.

Step 0 (trivial case). If ${\displaystyle X\in {\mathcal {A}}}$, then *every* ${\displaystyle \delta >0}$ works (because every ${\displaystyle B\subset X}$ is contained in ${\displaystyle X}$). So assume ${\displaystyle X\notin {\mathcal {A}}}$.

Step 1 (finite subcover). By compactness of ${\displaystyle X}$, choose a finite subcover

${\displaystyle X=\bigcup _{i=1}^{n}A_{i},\qquad A_{i}\in {\mathcal {A}}.}$

For each ${\displaystyle i}$, set the closed complement

${\displaystyle C_{i}=X\setminus A_{i}.}$

(Each ${\displaystyle C_{i}}$ is closed because ${\displaystyle A_{i}}$ is open.)

• • Step 2 (build a strictly positive continuous function).** Define

${\displaystyle f:X\longrightarrow \mathbb {R} ,\qquad f(x)={\frac {1}{n}}\sum _{i=1}^{n}d(x,C_{i}).}$

Each map ${\displaystyle x\mapsto d(x,C_{i})}$ is continuous by the 1-Lipschitz fact above (we did **not** use that ${\displaystyle C_{i}}$ is closed to get continuity), so ${\displaystyle f}$ is continuous as a finite average of continuous functions.

We now show ${\displaystyle f(x)>0}$ for **every** ${\displaystyle x\in X}$.

Fix ${\displaystyle x\in X}$. Since ${\displaystyle \{A_{i}\}}$ covers ${\displaystyle X}$, choose ${\displaystyle i}$ with ${\displaystyle x\in A_{i}}$. Because ${\displaystyle A_{i}}$ is open, there exists ${\displaystyle \varepsilon >0}$ such that

${\displaystyle B(x,\varepsilon )\subset A_{i}.}$

We claim ${\displaystyle d(x,C_{i})\geq \varepsilon }$. Indeed, if some ${\displaystyle c\in C_{i}}$ satisfied ${\displaystyle d(x,c)<\varepsilon }$, then ${\displaystyle c\in B(x,\varepsilon )\subset A_{i}}$, contradicting ${\displaystyle c\in C_{i}=X\setminus A_{i}}$. Thus every ${\displaystyle c\in C_{i}}$ has ${\displaystyle d(x,c)\geq \varepsilon }$, hence

${\displaystyle d(x,C_{i})=\inf _{c\in C_{i}}d(x,c)\;\geq \;\varepsilon .}$

Therefore

${\displaystyle f(x)={\frac {1}{n}}\sum _{j=1}^{n}d(x,C_{j})\ \geq \ {\frac {1}{n}}\,d(x,C_{i})\ \geq \ {\frac {\varepsilon }{n}}\;>\;0.}$

Since ${\displaystyle x}$ was arbitrary, ${\displaystyle f(x)>0}$ on ${\displaystyle X}$.

Step 3 (take a positive minimum). Compactness of ${\displaystyle X}$ and continuity of ${\displaystyle f}$ imply that ${\displaystyle f}$ attains a minimum:

${\displaystyle \delta \;=\;\min _{x\in X}f(x)\quad {\text{exists.}}}$

By Step 2, ${\displaystyle f(x)>0}$ everywhere; hence the minimum ${\displaystyle \delta }$ satisfies ${\displaystyle \delta >0}$.

Step 4 (verify the Lebesgue property). Let ${\displaystyle B\subset X}$ have ${\displaystyle \operatorname {diam} (B)<\delta }$. Pick any ${\displaystyle x_{0}\in B}$. Then for every ${\displaystyle y\in B}$,

${\displaystyle d(y,x_{0})\ \leq \ \operatorname {diam} (B)\ <\ \delta .}$

Among the finitely many real numbers ${\displaystyle \{d(x_{0},C_{i})\}_{i=1}^{n}}$, choose an index ${\displaystyle m}$ realizing the maximum:

${\displaystyle d(x_{0},C_{m})\ =\ \max _{1\leq i\leq n}d(x_{0},C_{i})\ \geq \ {\frac {1}{n}}\sum _{i=1}^{n}d(x_{0},C_{i})\ =\ f(x_{0})\ \geq \ \delta .}$

(Here we used the basic inequality ${\displaystyle \max\{r_{i}\}\geq {\frac {1}{n}}\sum r_{i}}$ for finitely many real numbers.)

We claim ${\displaystyle B\subset A_{m}}$. It suffices to show ${\displaystyle B(x_{0},\delta )\subset A_{m}}$ because ${\displaystyle B\subset B(x_{0},\delta )}$ by the diameter bound.

So let ${\displaystyle y\in B(x_{0},\delta )}$. Suppose, aiming for a contradiction, that ${\displaystyle y\in C_{m}=X\setminus A_{m}}$. Then, by definition of distance to a set,

${\displaystyle d(x_{0},C_{m})\ \leq \ d(x_{0},y)\ <\ \delta ,}$

contradicting ${\displaystyle d(x_{0},C_{m})\geq \delta }$. Hence no point of ${\displaystyle B(x_{0},\delta )}$ lies in ${\displaystyle C_{m}}$; equivalently,

${\displaystyle B(x_{0},\delta )\ \subset \ X\setminus C_{m}\ =\ A_{m}.}$

Since ${\displaystyle B\subset B(x_{0},\delta )}$, we conclude ${\displaystyle B\subset A_{m}\in {\mathcal {A}}}$, as required.

This proves that ${\displaystyle \delta }$ is a Lebesgue number for ${\displaystyle {\mathcal {A}}}$. ${\displaystyle \square }$

1. It is not true in general for a metric space that a closed and bounded set is compact. Take the following metric on a set X:

${\displaystyle d(x,y)={\begin{cases}1&{\rm {if}}\,x\neq y\\0&{\rm {if}}\,x=y\end{cases}}}$

a) Show that this is a metric

b) Which subspaces of X are compact

c) Show that if Y is a subspace of X and Y is compact, then Y is closed and bounded

d) Show that for any metric space, compact sets are always closed and bounded

e) Show that with this particular metric, closed and bounded sets need not be compact