Why does a first axiom space have to be $T_{1}$ in order for limit points to have a sequence converging to them?

My textbook General Topology by Pervin) says "If $x$ is a point and $E$ a subset of a $T_{1}$-space $X$ satisfying the first axiom of countability, then $x$ is a limit point of $E$ iff there exists a sequence of distinct points in $E$ converging to $X$"?

I don't understand why $X$ needs to be a $T_{1}$-space for this to be true.

Let $X$ be a first axiom topological space, and $E\subseteq X$. If a point $a\in X$ is a limit point of $E$, then there is a sequence in $E$ whose limit is $a$. Proof: let $B_{n}(a)$ be a monotone decreasing countable open base containing $a$. All these sets will contain at least one point from $E$. If we keep on choosing a countable number of points from each $B_{i}(a)\cap E$, we will generate one such sequence converging to $a$.

It is also clearly true that if there is a sequence in $E$, which converges to $a$, then $a$ is a limit point of $E$.

Nowehere in this argument was the fact that $E$ is a $T_{1}$-space is used.

I feel $X$ being a $T_{1}$-space only proves that $\bigcap_{i}B_{i}(a)=\{a\}$. This fact I feel is irrelevant in making the argument concerned.

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Your text claims something more than you prove, namely that there is a sequence of distinct points converging to $x$. – Miha Habič Jun 16 '13 at 10:18
I don't see why that should not be possible in my argument. – Ayush Khaitan Jun 16 '13 at 10:20
Suppose that any neighbourhood of $x$ contains some point $y\in E$ and that $\{x,y\}$ is open. You cannot possibly get an injective sequence to converge to $x$ in this case. – Miha Habič Jun 16 '13 at 10:31
We know $y\in E$. $\{x,y\}$ has to be equal to some $B_{i}(x)$, and all $B_{j}(x)=B_{i}(x)$ for $j>i, i,j\in\mathbb{N}$ (as $x$ is a limit point of $E$). By selecting $y$ from every $B_{j}$ for $j>i$, I'm stil creating a sequence converging to $x$!! This would not be true if the statement said "infinite distinct points". However, this is not what the statement says. – Ayush Khaitan Jun 16 '13 at 10:54
As I (and I think most people) understand the statement, it is asserting that in a first countable $T_1$ space, a point is a limit point of $E$ if there is an injective sequence in $E$ converging to it. I do agree that first countability suffices to get some (not necessarily injective) sequence. – Miha Habič Jun 16 '13 at 11:07

The key word here is "distinct". So let $X$ be a fixed $T_1$-space. Let $E \subset X$.

First note that a limit point of $E$ by the standard definition is a point $x$ such that every (open) neighbourhood $U$ of $x$ contains a point of $E$ different from $x$.

Now, for a $T_1$-space we can say something stronger: $x$ is a limit point of $E$ iff every (open) neighbourhood of $x$ contains infinitely many points from $E$.

Proof: if every neighbourhood contains infinitely many points of $E$, it will certainly contain one different from $x$, so one implication is trivial, so to see the other one: suppose a limit point $x$ of $E$ has an open neighbourhood $U$ such that $U \cap E$ is finite. Then $F = (U \cap E) \setminus \{x\}$ is also finite, and thus a closed set (here we use $T_1$: singletons and thus finite sets are closed), and so $X \setminus F$ is open, and so is $U' = U \cap (X \setminus F)$. As the latter set is thus an open neighbourhood of $x$ such that $U' \cap E \subset \{x\}$ (i.e. either empty or just $x$, depending on whether $x \in E$ or not), this contradicts that $x$ is a limit point of $E$.

This is the reason the fact is stated for $T_1$-spaces. In a non-$T_1$ space like Sierpinski space $\{0,1\}$ with open sets $\{\emptyset, \{0,1\},\{0\}\}$, which is $T_0$, the point $1$ is a limit point of the finite set $E = \{0\}$, but we certainly cannot find infinitely many different points from $E$ converging to $1$, even though the space is trivially first countable (and the constant sequence of $0$'s does converge to $1$).

Now if there is a sequence of distinct points from $E$ that converges to $x$, then clearly $x$ is a limit point of $E$ (every open neighbourhood of $x$ contains a tail of the sequence, and so even infinitely many different points of $E$, so we get the strong version of limit point for free). This holds without $T_1$ or first countability.

Now if $X$ is $T_1$ and also first countable, and if $x$ is a limit point, then let $U_n$ be a local base at $x$, and pick $x_1 \in E \cap U_1$. When we have chosen $x_1,\ldots,x_n$ (all distinct and from $E$) such that $x_i \in U_j$ for all $j \le i$ ( for $i=1 \ldots n$), then pick $x_{n+1} \in E \cap (U_1 \cap \ldots \cap U_n \cap U_{n+1})$, such that $x_{n+1}$ is distinct from all $x_1,\ldots,x_n$. This can be done as $U_1 \cap \ldots U_n \cap U_{n+1}$ is an open neighbourhood of $x$ and every one of them intersects $E$ in infinitely many different points by the strengthened version of limit point, which is valid in $T_1$-spaces. This defines a sequence by recursion, and it clearly converges to $x$ and consists of all distinct points from $E$.

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