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Let $X$ be a topological space that admits a countable open cover $\{U_i\}_i$ such that each $U_i$ is second countable in the subspace topology. Show that $X$ is second countable.

My attempted proof is as follows:

For each $i$, $U_i$ admits a countable basis $\mathcal{B}_i$. By the Axiom of Choice, $\cup_i \mathcal{B}_i$ is countable. All of the sets in $\cup_i\mathcal{B}_i$ are open in $X$ since they are of the form $B_i\cap U_i$ for some $B_i\in\mathcal{B}_i$, and both $B_i$ and $U_i$ are open. Since the $\{U_i\}$ are an open cover, for any $x\in X$, $x\in U_i$ for some $i$, so there is $B\in\mathcal{B}_i$ such that $x\in B$.

Now suppose that for any $B_1,B_2\in\cup_i\mathcal{B}_i$, $x\in B_1\cap B_2$. Then there are $i,j$ such that $B_1\in\mathcal{B}_i$ and $B_2\in\mathcal{B}_j$. But I do not understand how to show there is a $B_3\in\cup_i\mathcal{B}_i$ such that $x\in B_3\subset B_1\cap B_2$. How can we show there is such a $B_3$?

I know similar questions have been asked before but I couldn't find the answer to my question.

A. Goodier
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  • So your main question is "how can we show there is such a $B_3$"? Otherwise, it would be a possible duplicate of Topological space with countable open cover ${U_\alpha} $ with each $U_\alpha$ second-countable, is second countable – YuiTo Cheng Apr 16 '19 at 09:14
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    It's not hard to show that there is such a $B_3$ as you want, but why bother? To show that $\bigcup_i\mathcal B_i$ is a base for the topology of $X$, all you need to show is that, for any open set $U$ of $X$ and any point $x\in U$, there is a set $B\in\bigcup_i\mathcal B_i$ such that $x\in B\subseteq U$. That's the definition of what it means to be a base for a topology. – bof Apr 16 '19 at 09:14
  • The requirement that if $B_1,B_2$ are in the basis with nonempty intersection then there is a $B_3\subset B_1\cap B_2$ in the basis is part of the definition of basis that I am working with. – A. Goodier Apr 16 '19 at 09:17
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    I think you're confusing (1) the definition of what it means for a collection $\mathcal B$ to be a base for a given topology with (2) a theorem which gives you a necessary and sufficient condition for $\mathcal B$ to be a base for *some* topology. After you prove that your $B_3$ exists, all you know is that $\bigcup_i\mathcal B_i$ is a base for *some* topology, not necessarily the one you're interested in. – bof Apr 16 '19 at 09:18
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    @bof I think you're right. Thanks for clarifying. – A. Goodier Apr 16 '19 at 09:25
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    Instead of $\bigcup_i\mathcal B_i$ why don't you just use ${X}$? You will have no difficulty in showing that, if $B_1,B_2\in{X}$ and $x\in B_1\cap B_2$, then there is $B_3\in{X}$ such that $x\in B_3\subseteq B_1\cap B_2$. Therefore ${X}$ is a base for a topology on $X$, and that topology is obviously second countable. Of course it does not have anything to do with the given topology. – bof Apr 16 '19 at 09:27

3 Answers3

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You know that the intersection $B = B_1 \cap B_2$ is open in $X$. Since $B_1 \in \mathcal{B}_i$, we have $B \subset U_i$, i.e. $B$ is open in $U_i$. Now you find $B_3 \in \mathcal{B}_i \subset \bigcup_i \mathcal{B}_i$ such $x \in B_3 \subset U_i$.

Edit:

bof rightly remarked that you have to show something else: For each $x \in X$ and each open neigborhood $U$ of $x$ there exists $B \in \mathcal{B} = \bigcup_i \mathcal{B}_i$ such that $ x \in B \subset U$. This is done by the same argument as above: There exists $U_i$ such that $x \in U_i$. Then $U' = U \cap U_i$ is open in $U_i$ and you find $B \in \mathcal{B}_i \subset \mathcal{B}$ such $x \in B \subset U' \subset U$. In your question you only consider the special case $x \in U = B_1 \cap B_2$.

Paul Frost
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Hint: you are trying to prove something that is not necessary. Consider the collection sof all finite intersections of sets from $\cup_i \mathcal B_i$. This is a countable collection. Show that this is a basis for the topology of $X$.

Kavi Rama Murthy
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There exists $U$ such that $ x \in U$.

We know $B_1 \cap U$ and $B_2 \cap U$ are open in $X$, and both are subsets of $U$, and also their intersection $B'$ is open. Then there exists $B_3$ a subset of $B'$ containing $x$ by the property of basis.

with-forest
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