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Let $X$ be a metric space. Show that: If $X$ has a countable dense subset, then every open cover of $X$ has a countable subcover.

My attempt:Suppose $X$ has a countable dense subset, $D$. Let $(U_i)_{i\in I}$ be an open cover for $X$. Since $\overline{D}$= $X$ and $D$ is countable (enumerate its elements as $x_1,x_2.....$ ), then for each $x\in D$, there exists $U_i$, for which, there exists a radius $r_x>0$ such that $B(x,r)\subseteq U_i$. Put $S=\{$ $U\in (U_i)_i$ $:$ $\exists B(x_i,r_i)\subseteq U$ for some $x_i\in D$ $\}$. Then, since there are countably many balls , $S$ is countable. Therefore, it suffices to show that the members of $S$ form a cover for $X$. If $x\in X$ then there exists $x\in U \in (U_i)_{i\in I}$ such that $U\cap D\neq \varnothing$. Hence, there exists $z\in D$, for which, $z\in U$, since $U$ is open, there exists $r>0$, for which, $B(z,r)\subseteq U$.

Is this correct? (Please answer this question)

Brian M. Scott
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  • Unfortunately, your $S$, as you have defined it, need not be countable: it’s entirely possible that $A_1={i\in I:B(x_1,r_1)\subseteq U_i}$ is uncountable, for instance, and $U_i\in S$ for each $i\in A_1$. – Brian M. Scott Apr 19 '20 at 19:32
  • @Arturo Magidin I am so so sorry, it’s hell without my glasses. –  Apr 22 '20 at 01:14

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Your argument is not quite correct. The set $S$ you define is not necessarily countable. Consider the metric space $\mathbb{R}$ and the dense set $\mathbb{Q}$. Consider the open cover of all nonempty open sets. Then in your construction there is some $r_0$ corresponding to $0 \in \mathbb{Q}$. But there will be uncountably many open sets that contain $B(0, r_0)$, each of these will be in $S$, so $S$ will be uncountable.

If you want a proof, see this post.

Physical Mathematics
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