0.1  (§25) Components and local connectedness

• Def. $x\sim y$ iff some connected $C\subset X$ contains both. An equivalence relation; classes are the components of $X$.

• Thm. Components are connected, disjoint, and cover $X$; every nonempty connected subset lies in exactly one.

  Proof. Connectedness of a component $C$: fix $x_0\in C$, write $C={\bigcup }_{x\in C}{A}_x$ with each $A_x$ connected containing $x_0$; apply the common-point theorem.

• Thm. A finite product of connected spaces is connected.

  Proof. In $X\times Y$: $X\times \{y\}$ then $\{x^{\prime }\}\times Y$ links $(x,y)$ to $(x^{\prime },y^{\prime })$. Then induct.

  Rmk. True for arbitrary products too.

• Def. $x\simeq y$ iff a path joins them; classes are the path components. Same statements as for components.

• Ex. $\overline{S}$ (topologist’s sine curve): one component, two path components $S$ and $V$. Note $S$ is open but not closed in $\overline{S}$; $V$ is closed but not open.

• Def. $X$ is locally connected at $x$ if every neighborhood $U$ of $x$ contains a connected neighborhood $V$ of $x$: $x\in V\subset U$. Likewise locally path connected.

• Ex. $ℝ$ is locally (path) connected. $\overline{S}$ is not: small neighborhoods of points of $V$ are disconnected.

• Caution. Connected $\ne$ locally connected, in either direction.

• $\mathrm{★}$ Thm. $X$ locally connected $\Rightarrow$ every component is open. $X$ locally path connected $\Rightarrow$ every path component is open.

• $\mathrm{★}$ Thm. Each path component lies in a unique component. If $X$ is locally path connected, components $=$ path components.

  Proof. $C=P\bigsqcup U$ with $U$ a union of other path components; all are open, so $P$ is clopen in the connected $C$, forcing $P=C$.

0.2  (§26) Compact spaces

• Def. $𝒜=\{U_{\alpha }\}$ covers $X$ if $\bigcup U_{\alpha }=X$; an open cover if all $U_{\alpha }$ are open. A subcover is a subcollection that still covers.

• $\mathrm{★}$ Def. $X$ is compact if every open cover of $X$ has a finite subcover.

• Ex. Finite spaces are compact. $ℝ$ is not: ${\{(n-1,n+1)\}}_{n\in \mathbb{N}}$ has no finite subcover. But $ℝ$ with ${𝒯}_{\mathrm{triv}}$ is compact.

  Note. Compactness is a property of the topology, not of the set — worth saying out loud before students attach it to “bounded”.

• Rmk. Story break. Tell the Atiyah–Segal examination anecdote from S. G. Krantz, Mathematical Apocrypha: a nervous student, asked for an example of a compact set, answers “the real line” — and Segal rescues him with “in what topology?” The joke is exactly the previous two examples.

• Ex. $X=\{0\}\cup \{1/n\mid n\in \mathbb{N}\}\subset ℝ$ is compact: one $U\ni 0$ catches all but finitely many $1/n$; cover those individually.

• Ex. Any $X$ with ${𝒯}_{\mathrm{cof}}$ is compact: one nonempty $U_{\beta }$ leaves a finite complement.

• $\mathrm{★}$ Lem. $Y\subset X$ is compact $\iff$ every cover of $Y$ by open subsets of $X$ has a finite subcollection covering $Y$.

  Proof. Translate along $V_{\alpha }=Y\cap U_{\alpha }$. Lets you work in $X$ and forget the subspace topology.

• $\mathrm{★}$ Thm. A closed subspace of a compact space is compact.

• $\mathrm{★}$ Thm. A compact subspace of a Hausdorff space is closed.

  Proof. Fix $p\notin K$. For each $q\in K$ separate $p,q$ by $U_q,V_q$. Finitely many $V_{q_i}$ cover $K$; then $U=\bigcap U_{q_i}$ misses $K$. Compact sets generalise finite sets here.

• Lem. (Extracted from that proof.) $X$ Hausdorff, $K$ compact, $p\notin K$ $\Rightarrow$ there are disjoint open $U\ni p$ and $V\supset K$.

• $\mathrm{★}$ Thm. The continuous image of a compact space is compact.

• $\mathrm{★}$ Thm. $f:X\to Y$ a map, $X$ compact, $Y$ Hausdorff. Then

  1. $f$ is a closed map;

  2. $f$ surjective $\Rightarrow$ $f$ is a quotient map;

  3. $f$ bijective $\Rightarrow$ $f$ is a homeomorphism;

  4. $f$ injective $\Rightarrow$ $f$ is an embedding.

  Proof. All from (1): $A$ closed $\Rightarrow$ $A$ compact $\Rightarrow$ $f(A)$ compact $\Rightarrow$ $f(A)$ closed. This is the standard device for upgrading continuous bijections — contrast $[0,1)\to S^1$.

• Ex. $ℝ\to \{0\}$: preimages of compact sets need not be compact. Def. $f$ is proper if $f^{-1}(K)$ is compact for every compact $K$.

• $\mathrm{★}$ Lem. (Tube lemma.) $Y$ compact, $N\subset X\times Y$ open with $\{p\}\times Y\subset N$ $\Rightarrow$ there is a neighborhood $U\ni p$ with $U\times Y\subset N$.

  Proof. Cover $\{p\}\times Y$ by basis boxes $U_q\times V_q\subset N$; finitely many $V_{q_i}$ suffice; take $U=\bigcap U_{q_i}$.

• Caution. Fails without compactness of $Y$: take .

• $\mathrm{★}$ Thm. $X,Y$ compact $\Rightarrow X\times Y$ compact. Cor. Finite products of compact spaces are compact.

  Proof. For each $p$, finitely many $W_{\alpha }$ cover $\{p\}\times Y$; the tube lemma widens this to $U_p\times Y$. Then finitely many $U_p$ cover $X$. Union of finitely many finite collections.

• Def. $𝒞$ has the finite intersection property if every finite subcollection has nonempty intersection.

• $\mathrm{★}$ Thm. $X$ is compact $\iff$ every collection $𝒞$ of closed sets with the finite intersection property has ${\bigcap }_{C\in 𝒞}C\ne \mathrm{\varnothing }$.

  Proof. Purely the contrapositive of the definition, read through complements. Write the four successive rephrasings on the board.