First-countable spaces, closure of a subset and limit of a function From Wikipedia:


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One of the most important properties of first-countable spaces is that
  given a subset $A$, a point $x$ lies in the closure of $A$ if and only
  if there exists a sequence $\{x_n\}$ in $A$ which converges to $x$.

I was wondering if the above quote is equivalent to that there is no
isolated point in a first-countable space?

*I was wondering when"$f$ is a function on a first-countable space"
as in the following quotes, what the codomain of $f$ is?

This has consequences for limits and continuity. 
In particular, if $f$ is a function on a first-countable space, then
  $f$ has a limit $L$ at the point $x$ if and only if for every sequence
  $x_n → x$, where $x_n ≠ x$ for all $n$, we have $f(x_n) → L$. 
Also, if $f$ is a function on a first-countable space, then $f$ is
  continuous if and only if whenever $x_n → x$, then $f(x_n) → f(x)$.

Thanks and regards!
 A: (1) No, it is not. Indeed, a discrete space, in which every point is isolated, is automatically first countable: every point has a local base of cardinality $1$. I suspect that you’re overlooking the possibility of constant sequences: an isolated point is in the closure of a set $A$ iff it is already in $A$, in which case it is the limit of a constant sequence in $A$.
(2) The codomain doesn’t make any difference: the assertion is true irrespective of the codomain.

Proof: Suppose first that $X$ is first countable and $f:X\to Y$ is continuous. Let $x\in X$, and let $\langle x_n:n\in\omega\rangle$ be a sequence in $X$ converging to $x$. Let $U$ be any open nbhd of $f(x)$ in $Y$. Then $V=f^{-1}[U]$ is an open nbhd of $x$ in $X$, so there is an $m\in\omega$ such that $x_n\in V$ whenever $n\ge m$. But then for every $n\ge m$ we have $f(x_n)\in U$. Thus, each open nbhd of $f(x)$ contains a tail of $\langle f(x_n):n\in\omega\rangle$, which therefore converges to $f(x)$.
Now suppose that $f$ is not continuous. Then there is an open set $U$ in $Y$ such that $V=f^{-1}[U]$ is not open in $X$. Since $V$ is not open, there must be an $x\in V$ that is not in the interior of $V$. Let $\{B_n:n\in\omega\}$ be a countable local base at $x$, and assume without loss of generality that $B_0\supseteq B_1\supseteq\dots$. Since $x$ is not in the interior of $V$, for each $n\in\omega$ there is a point $x_n\in B_n\setminus V$, and it’s an easy exercise to check that $\langle x_n:n\in\omega\rangle$ converges to $x$. But for every $n\in\omega$, $f(x_n)\in Y\setminus U$, so $\langle f(x_n):n\in\omega\rangle$ does not converge to $f(x)$. $\dashv$

