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Is this proof correct?

Claim: if $(M,d)$ is a separable metric space, then so is any subset $U\subseteq M$

Proof:

  • let $C=(c_k)_{k\in \mathbb{N}}$ be a countable dense subset of $M$. Fix $n \in \mathbb{N}$. Consider $B_{1/n}(c_k)$ (open ball around $c_k$ of radius $1/n$). If $B_{1/n}(c_k) \cap U$ is non empty, then add to the set $U_n$ a point $u_k^{n}$ in $B_{1/n}(c_k) \cap U$
  • every $U_n$ is non empty by density of $C$, and being every $U_n$ at most countable, so is $U:=\bigcup_n U_n$, which I now claim to be dense in $U$
  • let $u \in U$. Find $c_l\rightarrow_l u$ such that $d(c_l,u)< 1/l$. Thus $B_{1/l}(c_l)\cap U \neq \emptyset$, and by construction there is $u_l^l \in U_l\subseteq U$ with $d(u_l^l,c_l)<1/l$
  • by triangle inequality: $d(u_l^l,u)<1/(2l)$, so that $u_l^l\rightarrow_l u$
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  • $\begingroup$ In this answer I show that a separable metric space has a countable base. Then $U$ also has a countable base and we pick a point in each member of a countable base to get a dense subset again. This is in essence your argument too. $\endgroup$ – Henno Brandsma Feb 17 at 22:32
  • $\begingroup$ The proof would be smoother if your definition of dense were not based off sequences, but just used: $D$ is dense iff every ball $B(x,r)$ intersects $D$. $\endgroup$ – Henno Brandsma Feb 17 at 22:42
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While you said enough that I can guess with some confidence what you intended, you never really defined the sets $U_n$. Here’s what I think you had in mind:

Fix an arbitrary point $u\in U$. For all $n,k\in\Bbb Z^+$, if $U\cap B_{1/n}(c_k)\ne\varnothing$ let $u_k^n\in U\cap B_{1/n}(c_k)$, and otherwise let $u_k^n=u$. Then for each $n\in\Bbb Z^+$ let $U_n=\{u_k^n:k\in\Bbb Z^+\}$.

(The only reason for $u$ is to allow me to define $u_k^n$ for all $n,k\in\Bbb Z^+$; this makes the later definition of $U_n$ a little simpler.)

It is clear that each $U_n$ is at most countable, and my small trick wth $u$ ensures that each $U_n$ is non-empty. If you define $u_k^n$ only when $U\cap B_{1/n}(c_k)\ne\varnothing$, then you actually have to do a bit of work to show that each $U_n$ is non-empty. You can argue that the fact that $C$ is dense in $M$ ensures that $B_{1/n}(u)\cap C\ne\varnothing$ for each $n\in\Bbb Z^+$, so for each $n\in\Bbb Z^+$ there is a $c_{k(n)}\in C\cap B_{1/n}(u)$. But then $u\in U\cap B_{1/n}(c_{k(n)})$, so $u_{k(n)}^n\in U_n$. (Here $u$ is any fixed point of $U$.) Thus, it is true that $\bigcup_{n\in\Bbb Z^+}U_n$ is a countable subset of $U$, but you cannot name it $U$: in your second bullet point you are using $U$ as a name for two different sets. I’ll call it $D$: $D=\bigcup_{n\in\Bbb Z^+}U_n$.

Your argument to show that $D$ is dense in $U$ would be right if you hadn’t got some of the notation wrong. Let $u\in U$; there is indeed a sequence in $C$ converging to $u$, but you’ve already indexed $C$ as $C=\{c_\ell:\ell\in\Bbb Z^+\}$, so you can’t just say that there is a sequence $\langle c_\ell:\ell\in\Bbb Z^+\rangle$ in $C$ that converges to $u$: you then have two competing (and almost certainly conflicting) indexings of $C$.

Use a different name: there is a sequence $\langle x_\ell:\ell\in\Bbb Z^+\rangle$ in $C$ that converges to $u$ and moreover has the property that $d(u,x_\ell)<\frac1\ell$ for each $\ell\in\Bbb Z^+$. For each $\ell\in\Bbb Z^+$ there is a $k(\ell)\in\Bbb Z^+$ such that $x_\ell=c_{k(\ell)}$, so $u\in U\cap B_{1/\ell}(c_{k(\ell)})$, and therefore $u_{k(\ell)}^\ell\in U\cap B_{1/\ell}(c_{k(\ell)})=U\cap B_{1/\ell}(x_\ell)$. Thus, $d(u_{k(\ell)}^\ell,x_\ell)<\frac1\ell$, and the triangle inequality ensures that

$$d(u,u_{k(\ell)}^\ell)\le d(u,x_\ell)+d(x_\ell,u_{k(\ell)}^\ell)<\frac2\ell\,.$$

(Note that the bound is $\frac2\ell$, not $\frac1{2\ell}$.) It follows that the sequence $\langle u_{k(\ell)}^\ell:\ell\in\Bbb Z^+\rangle$ in $D$ converges to $u$ and hence that $D$ is dense in $U$.

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  • $\begingroup$ Terrific answer! Thank YOU Prof. Scott! Your contribution is beyond words. We owe you so much $\endgroup$ – Theoneandonly Feb 19 at 2:05

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