Skip to main content

Topology MATH 342 Spring 2026

Kyle Ormsby

Section 2.5 Week 5

Subsection Monday

Subsubsection Compactness via topological induction

The following technique, topological induction, is an extremely useful proof strategy. In the near term, we will use it to prove that closed intervals of the form \([a,b]\subseteq \mathbb R\) are compact. Later, we will use the same method to show \([a,b]\) is connected, and it can also be used to prove existence and uniqueness of homotopy lifts for covering spaces.
🔗
Proof.
It should be clear to the reader that \([0,1]\) satisfies the given conditions. For the converse, suppose that \(C\subseteq [0,1]\) satisfies the conditions.
🔗
By the first condition, \(C\) is nonempty. Since \(C\subseteq [0,1]\text{,}\) we know that \(C\) is bounded above by \(1\text{.}\) By the least upper bound property for \(\mathbb R\text{,}\) the supremum of \(C\) exists and is in \([0,1]\text{.}\)
🔗
Since \(C\) is downwards closed, we must have \(C = [0,\sup C)\) or \(C = [0,\sup C]\text{.}\) But \(C\) is supremal, so \(C = [0,\sup C]\text{.}\)
🔗
Finally, the topological induction property ensures that \(\sup C=1\text{,}\) whence \(C = [0,1]\text{.}\)
🔗
🔗
We now provide the promised application of topological induction.
🔗
Proof.
If \(a=b\text{,}\) then \([a,b]=\{a\}\) is compact since \(a\) is necessarily an element of some set in an open cover, and thus is covered by a subcover consisting of a single set.
🔗
Henceforth, assume \(a\lt b\) and let \(\mathscr U = \{U_i\mid i\in I\}\) be an open cover of \([a,b]\text{.}\) The reader may check the following assertion:
Let \(x\in [a,b)\text{.}\) Then there is an element \(y\in (x,b]\) and an element \(i\in I\) for which \([x,y]\subseteq U_i\text{.}\)
🔗
🔗
🔗
Now let
\begin{equation*} C := \{y\in [a,b]\mid [a,y]\text{ can be finitely covered by elements of }\mathscr U\}. \end{equation*}
We employ topological induction to prove that \(C = [a,b]\text{,}\) whence \([a,b]\) is compact. The above assertion applied to \(x=a\) reveals that \(a\in C\text{.}\) Downward closure is similarly straightforward: if \(y\in C\) and \(a\le z\le y\text{,}\) then \(z\in C\) because a finite subcover of \([a,y]\) is automatically a finite subcover of \([a,z]\text{.}\)
🔗
For the topological induction step, suppose that \(y\in C\text{,}\) meaning that \([a,y]\) can be finitely covered by elements of \(\mathscr U\text{.}\) Fix a finite subcover of \([a,y]\text{.}\) Then \(y\) is in some \(U_i\) within the finite subcover. But \(U_i\) is open, so \(y+\varepsilon\) is in \(U_i\) for some \(\varepsilon \gt 0\text{.}\)
🔗
Our final task is to show that \(\sup C\in C\text{.}\) We already know that \(C = [a,\sup C)\) or \([a,\sup C]\text{.}\) Since \(\mathscr U\) is an open cover of \([a,b]\text{,}\) there is some index \(i_0\in I\) for which \(\sup C \in U_{i_0}\text{.}\) By openness of \(U_{i_0}\text{,}\) there exists \(z\in [a,\sup C)\) such that \((z,\sup C]\subseteq U_{i_0}\text{.}\) By definition of \(C\text{,}\) \([a,z]\) is finitely covered by \(\mathscr U\text{.}\) This means there is a finite subset \(S\subseteq I\) for which
\begin{equation*} [a,z]\subseteq \bigcup_{i\in S} U_i. \end{equation*}
Thus
\begin{equation*} [a,\sup C] = [a,z]\cup (z,\sup C]\subseteq \bigg(\bigcup_{i\in S}U_i\bigg)\cup U_{i_0}. \end{equation*}
In particular, \([a,\sup C]\) is finitely covered by \(\mathscr U\text{,}\) so \(\sup C\in C\text{.}\) By topological induction, \(C=[a,b]\text{,}\) which concludes our proof.
🔗
🔗
🔗

Subsubsection Inheritance of compactness

It can be challenging — or at least tedious — to directly demonstrate that a space is compact. Thankfully, there are a number of ways in which compactness can be inherited by subspaces of or constructions with compact spaces. We explore this phenomenon here.
🔗
Proof.
Let \(\mathscr U\subseteq \tau_X\) be an open cover of \(C\text{.}\) Define
\begin{equation*} \tilde{\mathscr U} := \mathscr U \cup \{X\smallsetminus C\} \end{equation*}
Since \(C\subseteq \bigcup_{U\in \mathscr U}U\text{,}\) \(\tilde{\mathscr U}\) covers \(X\text{.}\) And since \(C\) is closed, \(X\smallsetminus C\) is open, and thus \(\tilde{\mathscr U}\) is an open cover of \(X\text{.}\)
🔗
By hypothesis, \(X\) is compact there exist \(U_1,\ldots,U_n\in \tilde{\mathscr U}\) such that
\begin{equation*} X = U_1\cup \cdots \cup U_n\cup (X\smallsetminus C). \end{equation*}
It follows that
\begin{equation*} C\subseteq U_1\cup \cdots \cup U_n. \end{equation*}
As such, we have demonstrated that every open cover of \(C\) admits a finite subcover, i.e., \(C\) is compact.
🔗
🔗
Next, we will show that the continuous image of a compact set is compact. From analysis, we know that in metric topologies, compact is the same as closed and bounded, so this is a topological version of the extreme value theorem.
🔗
Proof.
Let \(\mathscr V\subseteq \tau_Y\) be an open cover of \(fK\text{.}\) We aim to show that \(\mathscr V\) admits a finite subcover. Because \(f\) is continuous, for each \(V\in \mathscr V\text{,}\) \(f^{-1}V\subseteq X\) is open. Since \(fK \subseteq \bigcup_{V\in \mathscr V}V\text{,}\) we know that
\begin{equation*} K\subseteq \bigcup_{V\in \mathscr V}f^{-1}V \end{equation*}
is an open cover of \(K\text{.}\)
🔗
By compactness of \(K\text{,}\) there exist \(V_1,\ldots,V_n\in \mathscr V\) such that
\begin{equation*} K\subseteq f^{-1}V_1\cup\cdots\cup f^{-1}V_n. \end{equation*}
Applying \(f\text{,}\) we see that
\begin{equation*} fK\subseteq V_1\cup\cdots\cup V_n, \end{equation*}
so \(\{V_1,\ldots,V_n\}\subseteq \mathscr V\) is a finite subcover.
🔗
🔗
Next up: finite products of compacts spaces are compact. But to get there, we will need a result called the Tube Lemma.
🔗
Proof.
By definition of the product topology, openness of \(W\) implies that there exist open sets \(x\in U_x\subseteq X\) and \(y\in V_y\subseteq Y\) for which \((x,y)\in U_x\times V_y\subseteq W\text{.}\) This implies that
\begin{equation*} \mathscr V := \{V_y\mid y\in Y\}\subseteq \tau_Y \end{equation*}
is an open cover of \(Y\text{.}\)
🔗
Since \(Y\) is compact, there exist \(y_1,\ldots,y_n\in Y\) such that
\begin{equation*} Y = V_{y_1}\cup\cdots\cup V_{y_n}. \end{equation*}
Take
\begin{equation*} U := U_{y_1}\cap\cdots\cap U_{y_n}. \end{equation*}
As a finite intersection of open sets, \(U\) is open, and since \(x\in U_x\) for all \(x\text{,}\) \(U\) is an open neighborhood of \(x\text{.}\) Finally,
\begin{equation*} U\times Y\subseteq (U_{y_1}\times V_{y_1})\times\cdots\times(U_{y_n}\times V_{y_n})\subseteq W. \end{equation*}
🔗
🔗
Proof.
By induction, it suffices to prove that \(X\times Y\) is compact when \(X,Y\) are compact.
🔗
Let \(\mathscr A\) be an open cover of \(X\times Y\text{.}\) For \(x\in X\text{,}\) consider the subspace \(\{x\}\times Y\subseteq X\times Y\text{.}\) Since \(Y\) is compact, so is \(\{x\}\times Y\cong Y\text{.}\) As such, there is a subcollection
\begin{equation*} \mathscr A_x = \{A_{x,1},\ldots,A_{x,n_x}\}\subseteq \mathscr A \end{equation*}
for which
\begin{equation*} \{x\}\times Y\subseteq A_{x,1}\cup\cdots\cup A_{x,n_x}. \end{equation*}
Applying the tube lemma to \(A_{x,1}\cup\cdots\cup A_{x,n_x}\) implies the existence of an open neighborhood \(U_x\subseteq X\) of \(x\) for which
\begin{equation*} U_x\times Y\subseteq A_{x,1}\cup\cdots\cup A_{x,n_x}. \end{equation*}
🔗
Behold: \(\{U_x\mid x\in X\}\) is an open cover of \(X\text{,}\) and \(X\) is compact, so there exist \(x_1,\ldots,x_k\in X\) such that \(U_{x_1}\cup\cdots\cup U_{x_k} = X\text{.}\) It follows that
\begin{equation*} \{A_{x_i,j}\mid 1\le i\le k, 1\le j\le n_{x_i}\}\subseteq \mathscr A \end{equation*}
is a finite subcover. This demonstrates that \(X\times Y\) is compact.
🔗
🔗
🔗
🔗

Subsection Wednesday

Subsubsection Separation of disjoint compact subspaces

Proof sketch.
The reader should fill in the details of the following sketch.
  • If \(K\) or \(L\) is empty, then one of \(U,V\) can be empty and the other equal to \(X\text{.}\) Now assume both \(K,L\) are nonempty.
    🔗
    🔗
  • Fix \(\ell\in L\text{.}\) Given \(k\in K\) separate \(k,\ell\) with open neighborhoods \(U_k,V_k\subseteq X\text{,}\) respectively, where \(U_k\cap V_k=\varnothing\text{.}\)
    🔗
    🔗
  • The set \(\{U_k\mid k\in K\}\) is an open cover of the compact space \(K\text{,}\) so there is a finite subset \(S_\ell\subseteq K\) such that
    \begin{equation*} \bigcup_{k\in S_\ell} U_k \supseteq K. \end{equation*}
    🔗
    🔗
  • Take
    \begin{equation*} U_\ell := \bigcup_{k\in S_\ell} U_k\quad\text{and}\quad V_\ell := \bigcap_{k\in S_\ell} V_k. \end{equation*}
    🔗
    🔗
  • Verify that \(V_\ell\) is an open neighborhood of \(\ell\text{,}\) and thus \(\{V_\ell\mid \ell\in \ell\}\) is an open cover of \(L\text{.}\)
    🔗
    🔗
  • By compactness of \(L\text{,}\) there is a finite subset \(T\subseteq L\) such that \(\bigcup_{\ell\in T} V_\ell\supseteq L\text{.}\)
    🔗
    🔗
  • Finally, check that
    \begin{equation*} U := \bigcap_{\ell\in T}U_\ell\quad\text{and}\quad V := \bigcup_{\ell\in T}V_\ell \end{equation*}
    form the desired separation of \(K,L\) by disjoint open sets.
    🔗
    🔗
🔗
🔗
🔗
Lemma 2.5.7 immediately proves Theorem 2.5.3 by taking \(D\) to be a singleton.
🔗
🔗

Subsection Friday

Class was canceled this day. The planned content will appear Monday of week 6.
🔗
🔗
🔗
PrevTopNext