# Hausdorff spaces

1. Jul 8, 2013

### R136a1

Hello everybody!

It is known that in Hausdorff spaces that every sequence converges to at most one point. I'm curious whether this characterizes Hausdorff spaces. If in a space, every sequence converges to at most one point, can one deduce Hausdorff?

2. Jul 8, 2013

### WannabeNewton

Not necessarily. Let $X$ be an uncountable set with the cocountable topology $\mathcal{T} = \{U\subseteq X:X\setminus U \text{ is countable}\}$. Assume there exist two distinct points $p,p'$ and two neighborhoods $U,U'$ of the two points respectively such that $U\cap U' = \varnothing$. Then $X\setminus (U\cap U') = (X\setminus U)\cup (X\setminus U') = X$. But $(X\setminus U)\cup (X\setminus U')$ is a finite union of countable sets which is countable whereas $X$ is uncountable thus we have a contradiction. Hence $X$ is not Hausdorff under the cocountable topology.

Now let $(x_i)$ be a sequence in $X$ that converges to $x \in X$ and let $S = \{x_i:x_i\neq x\}$. This set is countable so $U\setminus S$ must be a neighborhood of $x$ in $X$. Thus there exists some $n\in \mathbb{N}$ such that $i\geq n\Rightarrow x_i\in U$ but the only distinct element of the sequence that is in $U$ is $x$ so $x_i = x$ for all $i\geq n$ i.e. any convergent sequence in $X$ must be eventually constant under the cocountable topology. Hence limits of convergent sequences must be unique (the map prescribing the sequence must be well-defined).

3. Jul 8, 2013

### R136a1

Great! Thanks a lot! The example you gave is very interesting since it has the same convergent sequences as the discrete topology.