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Suppose $X$ is a topological space that is locally euclidean of dimension some $n \in \Bbb{N}$. Show that $X$ is first countable.

My attempt: Let $p\in X$ and $U$ a neighborhood of $p$. By assumption, there exists a neighborhood $U'$ of $p$ such that $U'$ is homeomorphic to a first countable space. Hence $U'$ is first countable. Since $U\cap U' \subseteq U'$. It follows that $U \cap U'$ is first countable. Hence each $p \in U\cap U'$ has a local basis, $\mathbb{B}_p$. Since $U\cap U'$ is open in $U'$ and $U'$ is open in $X$, each term in $\mathbb{B}_p$ is open in $X$. Hence $\mathbb{B}_p$ the required local basis.

Is my attempt correct? What would be a better proof?

Martin Sleziak
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  • Wouldn't this follow immediately from the fact that neighborhood of p is homeomorphic to some open set of $\mathbb{R}^n$? – HelloDarkness Dec 26 '19 at 13:48
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    That implies that the neighborhood is first countable. Not every neighborhood is necessarily homeomorphic to $\mathbb{R}^n$ –  Dec 26 '19 at 13:50
  • Sorry, I confused the concepts! – HelloDarkness Dec 26 '19 at 13:54
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    Note that I assume the following definition of locally Euclidean of dimension $n$: every $p \in X$ has an open neighbourhood that is homeomorphic to $\Bbb R^n$. – Henno Brandsma Dec 26 '19 at 13:57

1 Answers1

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Almost. You forgot to mention countability at a later point in the proof ("has a local basis" instead of "has a countable local basis"). You don't need to start with $U$, that only distracts.

Could again be more direct: let $p \in X$ and $U$ its locally Euclidean open neighbourhood, so there is a homeomorphism $h: U \to \Bbb R^n$. $\Bbb R^n$ is first countable and so $U$ is too. Hence $p$ has a countable local base $\Bbb B_p$ in $U$, but as $U$ is open, all members of this local base are open in $X$ too, and moreover form a local base at $p$ for $X$: if $O$ (open in $X$) contains $p$, there is a member $B \in \Bbb B_p$ with $B \subseteq O \cap U \subseteq O$ ($O \cap U$ is open in $U$ and we apply the fact that $\Bbb B_p$ is a local base for $U$). As $p$ is arbitrary, $X$ is first countable.

Martin Sleziak
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Henno Brandsma
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    @topologicalmagician the homeomorphic fact follows immediately from your earlier proved fact about open continuous images: $U$ is the image of $\Bbb R^n$ under the continuous open $h^{-1}$ etc. But you didn't touch on it in your attempt, so I figured you knew that already. – Henno Brandsma Dec 26 '19 at 13:55
  • yup! I'm trying to follow the advice you mentioned earlier. –  Dec 26 '19 at 13:58
  • Out of curiosity, where would the proof fail if I wanted to show that every locally Euclidean space is second countable? This statement doesn't seem to be true, but where would the above proof fail if I replaced first countable with second? –  Dec 26 '19 at 14:09
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    @topologicalmagician It would be false: let $X=\mathbb{R} \times D$ be the product of $\Bbb R$ with a discrete uncountable space. This is locally Euclidean but not second countable. BTW, connected counterexamples also exist: look up the long line. – Henno Brandsma Dec 26 '19 at 14:12
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    @topologicalmagician An uncountable union of countable local bases need not be countable. The union of local bases always is a base, but the cardinality... – Henno Brandsma Dec 26 '19 at 14:13