# First-order topologies, limit-points and convergent sequences

Let $(X,T)$ be a topological space, $D\subseteq X$ be a subset, and $x\in X$ have a countable neighbourhood basis $(U_k)$.

I want to know how I can prove the following statement:

If x is a limit-point of $D$, then there exists a sequence $(x_k)$ in $D\backslash\{x\}$ with $x_k\rightarrow x$.

In the book I am currently using for my studies the author proves this statement for all points of metric spaces $X$ with the metric topology, constructing a sequence $(x_k)$ by taking one Element of each $\mathbb{B}(x,\frac{1}{k})\cap D\neq\emptyset$, where the sequence of open Balls around $x$ with radius $\frac{1}{k}$, $(\mathbb{B}(x,\frac{1}{k}))$, is obviously a countable neighbourhood basis of $x$.

In a later comment he states that the statement I wrote above is true, arguing that in the prove that I described he only used the property that every point of x has a countable neighbourhood. However, the author also used the property that $\mathbb{B}(x,\frac{1}{k})\supset (\mathbb{B}(x,\frac{1}{k+1}))$, as you can see above, and since for any point with a countable neighbourhood basis there needn't be any neighbourhood basis $(U_n)$ with $U_n\supset U_{n+1}$ for every $n\in\mathbb{N}$, I am unable to construct a series to prove the statement.

I have only been able to prove the statement for Elements with finite neighbourhood bases: Let $x$ have a finite open nbhd basis $U_1,...U_n$. Let $x$ be a limit-point of $D$. Now we simply take an element $$y\in (D\backslash\{x\})\cap\bigcap^{n}_{k=1}U_k$$ and we find that for the constant function $(y)_{n\in\mathbb{N}}$ we get the limit $x$, since for any nbhd $U$ of X we have $(D\backslash\{x\})\cap\bigcap^{n}_{k=1}U_k\subset \bigcap^{n}_{k=1}U_k\subset U$.

Sadly, since only finite intersections of open sets are guaranteed to be open, this only works for the finite case. I would be very grateful if someone could at least give me an idea to how to solve this for countable infinite nbhd bases.

Thanks.

• @bof, Yes, I meant that in both cases. I have now edited the question. thanks – confusedStudent Jan 9 '18 at 0:49

Consider $V_{n}=U_{1}\cap\cdots\cap U_{n}$, $n=1,2,...$, then $V_{n+1}\subseteq V_{n}$ and pick up an $x_{n}\in V_{n}\cap D$ such that $x_{n}\ne x$. It is ready to see that $x_{n}\rightarrow x$.
• @confusedStudent note that the $V_n$ form a decreasing neighbourhood base at $x$. In general, if $x$ has a countable neighbourhood base it has a countable decreasing one. – Henno Brandsma Jan 9 '18 at 7:16