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Proposition: If metric space $M$ has a countable open base, then any open covering of $M$ admits a countable subcovering.

Definition: A collection of open subsets $\{U_i\}$ of $M$ is called an open base of $M$, if every of open set $O\subseteq M$ is expressible as $O=\bigcup_i U_i$.

Proof Attempt:

Let $\{\Omega_i\}_{i\in I}$, where $I$ is an arbitrary indexing set, be an open cover of $M$; that is $M \subseteq \bigcup_{i=1}^\infty \Omega_i$ and for all $i \in I$, $\Omega_i$ is an open set. First, we show that $\forall i\in I$, $\Omega_i \cap M$ is open relative to $M$. Let $x\in \Omega_i \cap M$, then there exists $\epsilon \gt 0$ such that $N_\epsilon (x)\subseteq \Omega_i$. But, if $N_\epsilon (x)\nsubseteq \Omega_i \cap M$, then it must be that there exists $n\in N_\epsilon (x)$ such that $n \notin M$. Relative to $M$, it does not affect whether $\Omega_i \cap M$ is open in $M$. What cannot be is $n \in M$ and $n \notin \Omega_i$. So, at any rate, $\Omega_i \cap M$ is open in $M$. Now, suppose $M$ has a countable open base $\{\mathscr{O}_k\}_{k=1}^\infty$. Thus, $\Omega_i \cap M=\bigcup_{k=1}^\infty \mathscr{O}_k$. Consider that $\{\Omega_i\cap M\cap\mathscr{O}_k\}_{k=1}^\infty$ is countable $\forall i \in I$, it covers $M$, and $\forall i\in I$, $\Omega_i\cap M\cap\mathscr{O}_k\subseteq\Omega_k$. Therefore, $\bigcup_{i=1}^\infty(\bigcup_{k=1}^\infty\Omega_i\cap M\cap\mathscr{O}_k)$ is a countable subcovering of an arbitrary open covering of $M$, because the union of countable set is countable.

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I noticed some inaccuracies in your attempt that you should think about:

  1. You 'waste' a lot of time proving things related to relative open vs. open. This is not necessary here, since $M$ is the only metric space you are considering and everything happens inside $M$. In particular $\Omega_i \cap M = \Omega_i$ etc.

  2. Also $$ \Omega_i \cap M=\bigcup_{k=1}^\infty \mathscr{O}_k $$ will not be correct in general, because the right side is equal to $M$.

  3. Moreover, $\{\Omega_i \cap M \cap \mathscr{O}_k\}_k$ does not cover $M$ but $\Omega_i$.

  4. You correctly noted in the beginning that $I$ is arbitrary, but in the end you index the $\Omega_i$ by $i \in \Bbb{N}$, in which case there was nothing to prove from the start.

  5. $\{\Omega_i \cap M \cap \mathscr{O}_k\}_{k,i}$ is not a subcover of $\{\Omega_i\}_i$ in general. Recall that a subcover of $\{\Omega_i\}_{i \in I}$ is of the form $\{\Omega_{l}\}_{l \in L}$ for some $L \subseteq I$.

A better way to proceed is to prove the following statements:

  1. Every $\Omega_i$ is the union of some of the $\{\mathscr{O}_k\}_k$, say $\Omega_i = \bigcup_{k \in J_i}\mathscr{O}_k$ where $J_i \subseteq \Bbb{N}$.
  2. The set $J = \bigcup_{i \in I} J_i$ is countable.
  3. The set $\{\Omega_{k} \:|\: k \in J\}$ is a countable cover of $M$.
  4. For each $k \in J$ there exists $i_k \in I$ such that $\mathscr{O}_{k} \subseteq \Omega_{i_k}$. Choose one $i_k$ for each $k \in J$.
  5. The set $\{\Omega_{i_k} \:|\: k \in J\}$ is a countable subcover of $\{\Omega_i\}_i$.
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  • $\begingroup$ at (1), I thought about assuming everything happens in $M$, but decided not to. at (2),(3),(4),(5).. I don't know what I was thinking. Thanks for pointing it out. I'll try the outline you provided. $\endgroup$ – TheLast Cipher Aug 4 '18 at 8:30

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