# First-countable spaces, closure of a subset and limit of a function

From Wikipedia

1. 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?

2. 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!

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(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 $f(x)$. $\dashv$

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+1, Thanks! About (1), I was wondering if it is true for a general topological space that "a point $x$ lies in the closure of $A$ if and only if there exists a sequence $\{x_n\}$ in $A$ which converges to $x$"? In other words, is "in a first countable space" really a sufficient condition for the previous statement to be true? – Tim Jan 31 '12 at 1:14
@Tim: No, if the space is not first countable, a point can be in the closure of a set without being the limit of a sequence in that set. I’ll give you an example in minute. – Brian M. Scott Jan 31 '12 at 1:16
Let $X$ be an uncountable set, and let $p\in X$. Say that $V\subseteq X$ is open iff either $p\notin V$, or $X\setminus V$ is countable. Every point of $X$ except $p$ is isolated, but $p$ has no countable local base. If $\langle x_n:n\in\omega\rangle$ is any sequence in $X\setminus\{p\}$, $X\setminus\{x_n:n\in\omega\}$ is an open nbhd of $p$ that contains no point of the sequence, yet $p$ is clearly in the closure of $X\setminus\{p\}$. – Brian M. Scott Jan 31 '12 at 1:20
Thanks! So the closure of a subset can be partitioned into the set of isolated points of the subset, the set of limit points of the subset, and the set of points that are neither of the previous two cases? Is there a concept for the third type of points in the closure of a subset? – Tim Jan 31 '12 at 2:49
@Tim: The three-way partition is correct. I’m not aware of any special name for the third kind, which are simply the points in $\operatorname{cl}A\setminus A$; the second and third together, of course, are the cluster points of $A$. – Brian M. Scott Jan 31 '12 at 2:50