# Friday, August 19

:::{.remark}
Today: dimensions are $d$, since we'll want to reserve $n$ to index countable sequences.
Recall a manifold is Hausdorff, second countable (exists a countable family $\ts{B_n}$ where $x\in U \subseteq M\implies \exists n$ with $x\in B_n \subset U$), and locally homeomorphic to $\RR^d$.
Reformulating this: $M$ has an **atlas**, i.e. homeomorphisms $\ts{\phi_\alpha: U_\alpha \to \phi_\alpha(U_\alpha) \subseteq \RR^d}$ where $M = \Union_{\alpha} U_\alpha$.
Note that we only require existence of an atlas, i.e. it is not part of the data.
These three properties imply others: normal, metrizable, paracompact, regular, etc.
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:::{.example title="?"}
Start with $d=0$, noting $\RR^0 = \ts{0} \neq \emptyset$.
This forces $U_\alpha = \ts{x_\alpha}$ to be a singleton for every $\alpha$, since the $\phi_\alpha$ are in particular set-theoretic bijections.
Since these must be open, $M$ must carry the discrete topology since points are open.

Any discrete space is Hausdorff, and note that if $\ts{B_ \alpha}_{\alpha\in A}$ is a base then $\size A \geq \size M$ since each $x\in B_{ \alpha} \subseteq \ts{x}$ forces $B_\alpha = \ts{x}$.
So if $M$ is to be second-countable, $M$ must be a countable set.
Thus dimension 0 topological manifolds are precisely countable discrete sets, and homeomorphisms are precisely bijections between sets of the same cardinality.
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:::{.remark}
If $M$ is second countable, it must have countably many connected components since each component is an open set.
It is harder to give examples of *connected* Hausdorff spaces what are locally homeomorphic to $\RR^d$ but not second countable -- see the long line.
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:::{.proposition title="?"}
Second countable spaces are **Lindelof**, i.e. every open cover admits a *countable* subcover.
Hence a manifold always has a countable atlas.
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:::{.proof title="?"}
Let $\ts{B_n}_{n=1}^\infty$ be a countable base and $\ts{V_\alpha}$ be an open cover.
Consider only those $B_n$ which are contained in some $V_\alpha$, which exist by the definition of a base, and let $S\da\ts{n\st \exists \alpha, B_n \subseteq V_{\alpha}}$.


:::{.claim}
Claim: $\Union_{n\in S} B_n = M$.
Proof: $x\in M\implies x\in V_{\alpha}$ for some $\alpha$, and since $\ts{B_n}$ is a base there is some $N$ with $x\in B_n \subseteq V_{\alpha}$ and moreover $N\in S$ by definition.
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Now given $x\in M$, for all $n\in S$ choose $\alpha_n$ with $B_n \subseteq V_{\alpha_n }$; then $M = \Union_{n\in S} V_{\alpha_n}$ and this yields a countable subcover.
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