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I want to show the title. More formally, Let $X\subset\Bbb R^n$ be a subset and $\{U_\alpha\}_{\alpha\in I}$ be a open covering of $X$. Then there is a subcover $\{U_\alpha\}_{\alpha\in I'\subset I}$ that covers $X$.

My basic plan is that define $S:=\Bbb Q^n$. Then for each $r\in S$, if there is $U_r\in\{U_\alpha\}_{\alpha\in I}$ that contains $r$ then choose only one such $U_r$. Collect them all for each $r$ (if there's such). I want to show this $\{U_r\}$ covers $X$. Any help? or any other idea?

one potato two potato
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2 Answers2

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Your idea will, unfortunately, not quite work. Take for instance $X = \mathbb R$. Enumerate the rationals $\mathbb Q$ as $q_n$ and let $U_n = (q_n - 2^{-n}, q_n + 2^{-n})$. Take also $U_* = \mathbb R$ itself. Now, your construction is to take exactly one open neighborhood of every rational point. This won't work for any arbitrary choice. For instance, each $q_n \in U_n$ but the union $\bigcup U_n$ is not all of $\mathbb R$. The sum of the lengths of all the $U_n$ is $\sum 2^{-n} = 2$, so they cannot cover $\mathbb R$ even though we have a neighborhood of every rational point.

But as I said, your intuition to use $\mathbb Q$ is good. Recall that the definition of an open subset $U$ of $\mathbb R$ is that for all $x \in U$ there is a ball containing $x$ that is in $U$. Symbolically, $x \in B \subseteq U$. Such a ball is determined by two parameters - its center $p \in \mathbb R^n$ and its radius $r > 0$, denoted $B = B_p(r)$. We can perturb this a bit to show that $U$ is open iff every point $x \in U$ is contained in a rational ball $B \subseteq U$. We say that $B = B_p(r)$ is rational if $p \in \mathbb Q^n$ and $r \in \mathbb Q$. Importantly, there are only countably many rational balls!

Here's the proof. If $U$ is covered by rational balls then it's in particular covered by balls and is therefore open. Suppose conversely that $U$ is open and take $x \in U$. Then by assumption there is a ball $x \in B_p(r) \subseteq U$. By density, we can find rational points arbitrarily close to $x$. Then indeed we can find a rational point $q \in \mathbb Q^n$ such that $x \in B_q(\varepsilon) \subseteq B_p(r)$ Imagine this as nudging $p$ ever so slightly to a very nearby rational point $q$. We can additionally find some rational $s > \varepsilon$ that is just ever so slightly larger, so that $B_q(s) \subseteq B_p(r)$. Then $x \in B_q(s) \subseteq U$ is a rational ball. I didn't do this in great detail, but if you draw a picture you should be able to prove this more rigorously.

Anyways, take now your open cover $\{U_\alpha\}_{\alpha \in I}$. Then for each $x \in X$ there is some rational ball $x \in U_{\alpha}$. By the above, we can find a rational ball $x \in B^\alpha \subseteq U_{\alpha}$. There are only countably many rational balls $B^\alpha$, and for each ball pick exactly one $U_\alpha$ containing it. Then we have chosen countably many elements of the cover. Furthermore, we have that each $x$ is in such a rational ball. Hence, these chosen $U_\alpha$ are a countable subcover.

paul blart math cop
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  • @Koro we don't need $x$ to be a rational point to find a rational ball around this - rational balls are a base for the topology. Also, I don't see why we need the $U_r$ to be disjoint? There are countably many balls which cover, and hence countably many open sets containing these balls. – paul blart math cop Mar 21 '21 at 04:53
  • In the last paragraph, you mean pick only one $U_\alpha$ that contains $B_q(s)$ for each $s$ and small $q$? – one potato two potato Mar 21 '21 at 05:14
  • @barista Yes, I could have worded it a bit better. Each $x$ is in some rational ball. There are countably many distinct rational balls, so for each distinct one pick a $U_\alpha$ containing it. – paul blart math cop Mar 21 '21 at 05:18
  • Ok, let me state in my word. I will denote the given collection $C$. So, for each $x\in X$, I can choose $x\in B_q(s)\subset U_\alpha\in C$ where $s\in S$ and $q = 1/n$ for some $n\in\Bbb N$. Now for each $B_{1/n}(s)$, choose only one $U_\alpha\in C$ if it exists and collection of all such $U_\alpha$ gives a subcollection $C'$ of $C$. If $x\in X$ then there is $x\in B_{1/n}(s)\subset U_\alpha\in C$ so there is $U_\beta\in C'$ that contains $B_{1/n}(s)$ so it covers $X$ – one potato two potato Mar 21 '21 at 05:27
  • @barista Yup, that's exactly right. – paul blart math cop Mar 21 '21 at 05:35
  • Thank you very much! – one potato two potato Mar 21 '21 at 05:37
  • @barista Glad to help! – paul blart math cop Mar 21 '21 at 05:44
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$\mathbb R^n $ is second countable: the open balls with centers with rational coordinates and rational radii form a basis for the topology.

The property you are referring to is called the Lindelöf property. Any second countable space is Lindelöf.