Showing that $F$ is closed, given that it contains all of its limit points Let $X$ be a Hausdorff topological space. I am trying to show that if $F \subset X$ has the property that if $\{x_n\}_{n \in \mathbb{N}}$ is a sequence in $F$ that converges to $x_0 \in X$, we must have $x_0 \in F$, then $F$ is closed. 
My idea was to show that $F$ is equal to it's closure (so it suffices to show that $\overline{F} \subset F$). So I take a point $x_0 \in \overline{F}$ and if I can find a sequence in $F$ that converges to it then it's in $F$. But I'm having a hard time getting my hands on a sequence. I tried using a sequence $\{y_n\}$ of points of $X$ and defining a sequence $\{x_n\}$ using the fact that $x_0 \in \overline{F}$, but I don't know how to "make the open neighborhoods shrink around the point" without a metric.
By the way, this is the definition of convergence I am using: $\{x_n\}$ converges to $x_0 \iff$ for every open neighborhood $U$ of $x_0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $x_n \in U$.
 A: Sequences don't make sense unless the space has a countable neighborhood base. But we can replace them by nets (or flters). With this change, we prove the claim: 
If $F$ is not closed, then $X\setminus F$ is not open. Then there is an $x\in X\setminus F$ such that if $U$ is open in $X$ and contains $x$, then $U\cap F\neq \emptyset$.
Let $\left \{ N_{\alpha } \right \}_{\alpha \in \Lambda}$ be a neighborhood base at $x$, which we use to turn $\Lambda $ into a directed set:
$\alpha _i\geq\alpha _j\Leftrightarrow N_{\alpha _i}\subseteq N_{\alpha _j}$.
Now since  $U_{\alpha }\cap F\neq \emptyset$, we may choose, for each $\alpha \in \Lambda$ an $f_{\alpha }\in N_{\alpha }\cap F$.
Define the net $f:\Lambda \rightarrow X$ by $\alpha \mapsto f_{\alpha }$.
By construction $f_{\alpha }\rightarrow x$. That is, for every neighborhood $U$ of $x,\ f_{\alpha }\ $is eventually in $U$, which gives us our contradiction since in this case by the statement of the exercise, we must have $x\in F$. 
A: The following works if the space is first countable (every point has a countable basis of (open) neighbourhoods). Also, I'm a bit rusty with this, please point out errors if you see any. 
A point $x$ is on the boundary of $F$ iff every neighbourhood of it has at least one point in $F$ and at least one point not in $F$.
Recall the following property: Let $B_1$ and $B_2$ be base elements. 
Then $\forall x \in B_1 \cap B_2$, $\exists B_3 \subset B_1 \cap B_2$ s.t. $x \in B_3$
This gives us a countable sequence of nested sets $B_i$, every one of which contains $x_0$, and, as each set is a neighbourhood of a point on the boundary of $F$, contain at least one point $x_i \in F$.
Then you have two possibilities:


*

*$x_i = x_0$ for some $i$, in which case you're done because that implies $x_0 \in F$, or,

*Every $B_i$ contains $x_i \neq x_0$ in $F$. As every neighbourhood of $x_0$ contains a basis set, it has at least one point in $F$ that is not $x_0$ itself. Thus, $x_0$ is a limit point of $F$.
Consequently, $F$ contains its boundary, and is closed.
A: What you are describing are sequential spaces -- spaces where convergence of sequences characterizes the topology. That's true in first countable spaces, hence in metric spaces, but not in general. A product of uncountably many copies of a sequential space (with ≥ 2 points) won't be sequential (basically because the union of countably many countable sets is countable). See this blog article by @DonMa -- https://dantopology.wordpress.com/tag/sequential-space/page/2/ for more on this. Worth reading too is Pete Clark's survey of convergence: http://alpha.math.uga.edu/~pete/convergence.pdf, which discusses sequences, nets and filters, two equivalent generalizations of sequences in terms of which topologies can be characterized.
