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A Hausdorff space is defined as one for which, for all $x_1 , x_2 \in X, x_1 \neq x_2,$ $\exists $ open neighbourhood of $x_1$, $U_1$, and of $x_2, U_2$ for which $U_1 \cap U_2 = \emptyset$.

It is easy to prove that limits of sequences are unique in a Hausdorff space, however I was not able to show that limits are unique only in a topological space that satisfies the Hausdorff condition. I have spent quite some time thinking about this, and would like to know if this is not the case (otherwise I will keep trying to construct a proof). So far, I have also looked online but no text I have found on Hausdorff spaces has said that limits are unique 'iff' a space is Hausdorff, from which I would infer that this is not the case.

EDIT: I have a feeling it has to do with whether a space is 'preregular' or not,, but I have only just encountered this concept and am trying to digest it still.

EDIT 2: I just discovered this post, which answers my question with the negative. The post can be closed.

Meep
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  • Do you really only want limits of sequences, or are you also interested in uniqueness of limits of filters (or, equivalently, nets)? Of course, the link in EDIT 2 answers the sequence version of the question. – GEdgar Sep 21 '18 at 14:16

1 Answers1

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The theorem is " Let $X$ be a first countable (which means $X$ has some countable base at any point of it) space. Then having the property any converges sequence has just one limit point in $X$ equivalents $X$ is Hausdorff space.

As you noticed opposite side is clear.

Proof for the first side: Suppose $X$ is not Hausdorff and having the unique limit property. There must two different distorting rules points of $X$ say $x, y$. Since X is first countable we have $\mathcal{B_x'}=\{B_n: n\in \mathbb{N}\}$ and $\mathcal{B_y'}=\{A_n: n\in \mathbb{N}\}$ which are countable base respectively x and y. Say $U_n=B_1\cap\ldots \cap B_n$ and $V_n=A_1\cap\ldots \cap A_n$ and so ${B_x}=\{U_n: n\in \mathbb{N}\}$, $\mathcal{B}_y=\{V_n: n\in \mathbb{N}\}$ are still bases at the points. Since $x$ and $y$ can not serated with opens, we can choose $x_n\in V_n\cap U_n$ for every $n$. One can easly see that the sequence $\{x_n: n\in \mathbb{N}\}$ converges the two points.

Further theorem is "X is Hausdorff iff every convergent net has a unique limit in X"

Henno Brandsma
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