# Sequences in Non-Hausdorff Spaces

I was told that in any space that is not Hausdorff there are at least two points such that any sequence converges to one iff it converges to the other. I don't know how to prove this. Could I have any help?

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A sequence converges to a point if it eventually enters and stays in every neighborhood of that point. –  Vectk Jul 7 '12 at 17:58
Not true as stated. Consider the space $\{x,y\}$, where $\emptyset, \{x\}, \{x,y\}$ are open. Then the sequence $y,y,y,...$ converges to $y$ and not $x$. –  Kevin Jul 7 '12 at 18:01
Use what @Brian said about sequences and then think about what it means for a space to Not be Hausdorff by negating the Hausdorff condition. –  Francis Adams Jul 7 '12 at 18:01

The statement "there are at least two points such that any sequence converges to one iff it converges to the other" is somewhat ambiguous.

It could be read as "there are $x, y$ with $x \ne y$ such that any sequence that converges to $x$ also converges to $y$". In this case it is equivalent to "there are $x, y$ with $x \ne y$ such that every neighbourhood of $y$ is also a neighbourhood of $x$", which is the negation of T1.

Alternatively, it could mean "there are $x, y$ with $x \ne y$ such that any sequence that converges to $x$ also converges to $y$ and any sequence that converges to $y$ also converges to $x$". In that case it is equivalent to "there are $x, y$ with $x \ne y$ such that every neighbourhood of $y$ is also a neighbourhood of $x$ and every neighbourhood of $x$ is also a neighbourhood of $y$", which is the negation of T0.

Either way, T2 does not come into it.

EDIT: Reading the question again, I noticed that I overlooked the second f in "iff". Taking that into account, the second interpretation is the right one.

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This is false, I think.

Let $X = \{a,b\}$ and define a topology on $X$ by delaring $U$ open iff $a\in U$ or $U = \emptyset$.

Then $X$ is not Hausdorff as $a$ and $b$ are two points which can not be separated by disjoint open sets (since every open set containing $b$ must contain $a$).

Then the constant sequence $b,b,b,b,...$ converges to $b$ but not $a$. To see this, notice that the sequence is not eventually in the open set $\{a\}$.

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This is not conclusive, as the constant sequence $a,a,\dots$ tends to both $a$ and $b$. Oh, I read the question wrong! This is indeed a counter example: –  Olivier Bégassat Jul 7 '12 at 18:02
Thank you. This is the second time my book was wrong. –  Parakee Jul 7 '12 at 18:03
So maybe to get this to be true you need the space to be $T_1$ non-Hausdorff. That would at least take care of this example. –  Francis Adams Jul 7 '12 at 18:05
@Francis: I'm not sure what happens in this case. I'll keep thinking about it. –  Jason DeVito Jul 7 '12 at 18:07

The statement is false even for $T_1$ spaces. Let $X$ be an uncountable set, and give $X$ the co-countable topology: $V\subseteq X$ is open iff either $X\setminus V$ is countable, or $V=\varnothing$. Then all countable subsets of $X$ are closed, to all finite subsets of $X$ are closed, and $X$ is therefore $T_1$. On the other hand, a sequence $\langle x_n:n\in\Bbb N\rangle$ is convergent iff it is eventually constant, i.e., iff there are an $x\in X$ and an $n_0\in\Bbb N$ such that $x_n=x$ for all $n\ge n_0$, and in that case it converges to $x$ and only to $x$.

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