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I know this is correct: if a topological space $X$ is first countable then for every $x\in \text{Cl}(A)$ where $A$ is a subset of $X$, there exists a sequence in $X$ which converges to $x$.

Then, what if $X$ is NOT first countable? Does anyone have an example of topological space which is NOT first countable but satisfies "for every $x\in \text{Cl}(A)$ where $A$ is a subset of $X$, there exists a sequence in $X$ which converges to $x$"?

Thanks in advance, I really appreciate any help.

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A topological space $X$ with the property that for every $A\subseteq X$ and $x\in\overline{A}$ there exists some sequence on $A$ that converges to $x$ is known as a Fréchet (Fréchet-Urysohn) space. A classical example of a Fréchet but not first countable topological space is the set $\mathbb{R}/\sim$ where $$x\sim y\text{ iff }x=y\text{ or }x,y\in\mathbb{Z}$$ endowed with the quotient topology.

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  • $\begingroup$ Thanks for the help! $\endgroup$ – lacm Jun 1 '17 at 6:25
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There's another good example of this phenomena considering the cofinite topology. Let $(X,\tau)$ be a noncountable topological space with the cofinite topology (For example, the real line). Since the set of finite sets of $X$ will also be noncuntable, using that show that $(X,\tau)$ is not first countable. Now, for the sequences take a look at this:

Claim: For every $x\in X$ and $(x_n)_{n\in\omega}$, the sequence converges to $x$ if and only if for every $y\neq x$ only finitely many elements of the sequence take the value $y$.

For the proof of this claim analyze convergence in the space $X$. Now, it'll be easy to prove that $X$ is a Fréchet-Urysohn space. However, this space is not that interesting because is not $T_2$.

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