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Suppose we work in a separable, complete metric space $X$. Let $Z$ be an uncountable subset of $X$, must there exist $x_0\in Z$ and a sequence $(x_n)_{n=1}^\infty$ of elements in $Z$ different from $x_0$ that converges to $x$?

This is true for the real line. How about in general? What if $X$ is a separable Banach space.

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  • $\begingroup$ Separable metric spaces are Lindelof. See Theorem 1 here. $\endgroup$ – David Mitra Jul 2 '16 at 16:56
  • $\begingroup$ In metric spaces, separability $=$ second countability. Second countability carries over to subspaces, hence so does separability. Now that countable dense subset of $Z$ works just fine. $\endgroup$ – Hmm. Jul 2 '16 at 17:00
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Yes, and we do not even need to assume completeness.

Suppose not, towards a contradiction. Then we can find open balls $U_z=B_{\epsilon_z}(z)$ around each $z\in Z$ such that $U_z\cap Z=\{z\}$. Let $q>0$ be such that uncountably many $z\in Z$ have $\epsilon_z>q$ (Exercise: why does such a $q$ exist?). Let $Z'=\{z\in Z: \epsilon_z>q\}$.

Now consider the smaller open balls $V_z=B_{q\over 3}(z)$ for $z\in Z'$. These are disjoint (Exercise: why?), and there are uncountably many of them since $Z'$ is uncountable. But this contradicts the assumption that the space was separable: If $D$ is any dense subset of $X$, $D$ must have at least one element $d_z$ in each $V_z$ for every $z\in Z'$. Since the $V_z$s are disjoint, $d_z=d_w\iff z=w$; but then $\vert D\vert\ge\vert Z'\vert$, and $Z'$ is uncountable.

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Let $D\subset X$ be countable and dense and $Z\subset X$. Assume there is no $z_0\in Z$ with a sequence $(z_n)_{n=1}^\infty$ of points in $Z$ converging to $z_0$. Then for every $z\in Z$ there exists $r=r(z)>0$ such that $B_r(z)\cap Z=\{z\}$. By decreasing $r(z)$ if necessary, we may assume that $r(z)=\frac1{n(z)}$ with $n(z)\in\Bbb N$. For $m\in \Bbb N$, let $Z_m=\{\,z\in Z\mid r(z)=\frac 1m\,\}$. Fix $m\in\Bbb N$. Then for each $z\in Z_m$, the open ball $B_{\frac1{2m}}(z)$ contains at least one element of $D$. On the other hand, these open balls are pairwise disjoint for different $z\in Z_m$. We conclude that $Z_m$ is countable. But then $Z=\bigcup_m Z_m$ is also countable!

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Let $d$ be a metric for $X.$ Let $ E$ be a dense subset of $X$.

For any $x\in X$ and $r>0,$ there exists $q\in (0,r/3)\cap \mathbb Q $ and $e\in E$ such that $d(x,e)<q.$ $$ \text { This gives }\quad x\in B_d(e,2q)\subset B_d(x,r).$$ (Because $y\in B_d(e,2q)\implies d(x,y)\leq d(x,e)+d(e,y)< q+2q<r$.)

Therefore $B=\{B_d(e,2q):q\in \mathbb Q^+\land e\in E\}$ is a base for $X.$ If $E$ is countable then so is $B.$

So a separable metrizable space $X$ is first-countable.(It has a countable base). And any subspace $Z\subset X$ must also be first-countable, so $Z$ cannot be an uncountable discrete subspace: If $Z$ is uncountable there must exist $z\in Z$ such that for every $n\in \mathbb N$ there exists $z_n\in Z$ with $0<d(z,z_n)<1/n.$

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