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My textbook (Principles of General Topology, by Pervin) says "A topological space $X$ satisfying the first axiom of countability is a Hausdorff space iff every convergent sequence has a unique limit."

Let $<x_{n}>$ be a sequence converging to $x\in X$. Let there be another point $y\in X$. By the first axiom of countability, for every open set $G$ containing $x$, there is a $B_{n}(x)\subseteq G$, and for every open set $H$ containing $y$, there is a $B_{n}(y)\subseteq H$. Also, for every open set containing $x$, for a certain N, all $x_{j}$, $j\geq N$ lie in that open set.

Let every open set containing $y$ intersect every open set containing $x$, and let the intersection consist of points other than points of the sequence $<x_{n}>$. Then, this is not a Hausdorff space, but the sequence $<x_{n}>$ still tends to the unique limit $x$ (provided, for every other point in $X$ apart from $y$, we assume there is at least one open set containing that point which does not intersect with an open set containing $x$. Hence the sequence can't have any of those points as its limit)!

Is this not a contradiction then? We have created a space satisfying the first axiom of countability, in which every sequence tends to a unique limit, but the space is not Hausdorff?

One possible contradiction I see is if we can define a sequence in the intersections between the open sets containing $x$ and $y$. All these intersections will intersect amongst themselves if ANY two open sets containing $x$ and $y$ intersect, and hence such a sequence may be possible. It would be great if someone could tell me whether this is the reason why the theorem stands true.

Thanks a lot!

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For reference purposes- first axiom of countability is for every point $x\in X$ there exists a countable family $\{B_{n}(x)\}$ such that for any open set $G$ containing $x$, there is a $B_{n}$ such that $B_{n}\subseteq G$. Any help would be greatly appreciated. – Ayush Khaitan Jun 11 '13 at 8:29
up vote 2 down vote accepted

It’s not true that every sequence in tends to a unique limit. Suppose that every open nbhd of $x$ intersects every open nbhd of $y$, let $\{B_n(x):n\in\Bbb N\}$ be a countable local base at $x$, and let $\{B_n(y):n\in\Bbb N\}$ be a countable local base at $y$. We may assume that these bases are nested, i.e., that $B_{n+1}(x)\subseteq B_n(x)$ and $B_{n+1}(y)\subseteq B_n(y)$ for each $n\in\Bbb N$.

For each $n\in\Bbb N$ we know that $B_n(x)\cap B_n(y)\ne\varnothing$, so there is a point $z_n\in B_n(x)\cap B_n(y)$; I claim that the sequence $\langle z_n:n\in\Bbb N\rangle$ converges to both $x$ and $y$. Let $U$ be any open nbhd of $x$; then there is an $m\in\Bbb N$ such that $B_m(x)\subseteq U$, and for each $n\ge m$ we then have

$$z_n\in B_n(x)\subseteq B_m(x)\subseteq U\;.$$

This shows that $\langle z_n:n\in\Bbb N\rangle$ converges to $x$. An exactly similar argument shows that it also converges to $y$. Thus, there is at least this one sequence in $X$ that converges to two different limits, and that’s all that we had to show.

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Brian, why may we assume these bases are nested? I seem to be a little confused with this. Isn't it perfectly possible they're not? Thanks – Ayush Khaitan Jun 11 '13 at 13:44
@Ayush: Yes, but if you’re handed a local base that isn’t nested, you can always convert it to one that is. If $\{U_n:n\in\Bbb N\}$ is any local base at $x$, let $$B_n(x)=\bigcap_{k\le n}U_k\;;$$ then $\{B_n(x):n\in\Bbb N\}$ is a nested base at $x$. – Brian M. Scott Jun 11 '13 at 18:11

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