# Any subset of separable metric space is separable

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 $$u_l^l\rightarrow_l u$$
• 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. – Henno Brandsma Feb 17 at 22:32
• 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 $D$. – Henno Brandsma Feb 17 at 22:42

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 $$U$$.

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