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Prove that topological space $X$ is Hausdorff if and only if limits of all nets in it are unique.

Let $X$ be Hausdorff and $S:D\to X$ be a net in it where $D$ is a directed set. Let $S$ has two limits $x,y\in X$. We need to prove that $x=y$. Suppose $x\neq y$, therefore by Hausdorffness, there exists $U,V$ open sets in $X$ containing $x,y$ respectively such that $U\cap V=\emptyset$. Now by defintion of convergence, there exist $m_1,m_2\in D$ such that for all $n\in D$, $n\geq m_1$ implies $S(n)\in U$ and $n\geq m_2$ implies $S(n)\in V$. Again by definition of directed set, there exists $n\in D$ such that $n\geq m_1$ and $n\geq m_2$. This implies $S(n)\in U\cap V$, a contradiction. Hence $x=y$ and our first part is proved.

Now for the converse, if we take limits of all nets are equal, and WLOG assume $X$ is not Hausdorff. Then for $x,y\in X$, $x\neq y$, we would have two open sets containing $x,y$ respectively which are not disjoint. But I can't get a contradiction this way. How to proceed for this converse part? Any help is appreciated.

Henno Brandsma
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am_11235...
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  • You just have to construct a net which converges to two at least two limits. Hint: even a sequence works. – Keen-ameteur Mar 08 '20 at 06:58
  • I don't see what's stopping you from reversing everything in the first part. Besides, you're attempt at the converse is misdirected. If $X$ is not Hausdorff, then any open sets containing $x$ and $y$ respectively intersect nontrivially, for some $x\ne y$. But, from this it becomes possible to get a net that converges to both $x$ and $y$. So your logic broke down a bit. –  Mar 08 '20 at 07:10
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    @Keen-amateur I don't think you can necessarily find a sequence converging to both $x$ and $y$. But you can find a net. Nets are more general. If the space were first countable, then you could get a sequence. Check me on this. –  Mar 08 '20 at 07:48
  • Wouldn't the sequence $z_{2n}=x$ and $z_{2n+1}=y$ work? – Keen-ameteur Mar 08 '20 at 08:02
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    @Keen-ameteur Not necessarily. You're banking on every open set that contains $x$ to also contain $y$, and vice-versa. This is a lot stronger than a failure in Hausdorffness. – user757704 Mar 08 '20 at 08:29
  • @user757704 I'm actually banking on just direction of what you stated. In essence if we assume WLOG that $x$ is contained in every open set containing $y$, then every sequence\net which has $y$ as limit also has $x$ as a limit. Or at least I think so. – Keen-ameteur Mar 08 '20 at 09:05
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    @Keen-ameteur Smallest possible example: $X = {1, 2, 3}$, and $\tau = {\emptyset, X, {1, 2}, {2, 3}, {2}}$. Note that $(X, \tau)$ is a topological space, that $1$ and $3$ cannot be separated by disjoint open sets, but the sequence $1, 3, 1, 3, 1, 3, \ldots$ does not converge to $1$ or $3$. Specifically, if we take the open neighbourhood ${1, 2}$ of $1$, then the sequence is not eventually inside this neighbourhood. (On the other hand, the constant sequence $2$ converges to both $1$ and $3$, as well as $2$.) – user757704 Mar 08 '20 at 10:58
  • Okay, thanks for that clarification. Perhaps what I had in mind was the negation of $X$ being $T_1$. – Keen-ameteur Mar 08 '20 at 13:02

1 Answers1

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Suppose $X$ is not Hausdorff. Then there exist some distinct $x, y \in X$ such that, for all open neighbourhoods $U$ and $V$ of $x$ and $y$ respectively, $U \cap V \neq \emptyset$.

Define the directed set $$A = \{(U, V) : \text{$U, V$ are open and } x \in U, y\in V\},$$ with the partial order $$(U_1, V_1) \le (U_2, V_2) \iff U_1 \supseteq U_2 \text{ and } V_1 \supseteq V_2.$$ Given $(U_1, V_1), (U_2, V_2) \in A$, we can define an upper bound $$(U, V) = \left(U_1 \cap U_2, V_1 \cap V_2\right).$$ Choose (with axiom of choice) a net $(x_{(U, V)})_{(U, V) \in A}$ such that $$x_{(U, V)} \in U \cap V.$$ I claim that $x_{(U, V)} \to x, y$.

For any open neighbourhood $U_0$ of $x$, we have $$(U, V) \ge (U_0, X) \implies x_{(U, V)} \in U \cap V \subseteq U \subseteq U_0,$$ hence $x_{(U, V)} \to x$. A similar argument works for $y$ too.

user757704
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  • I presume you can define the upper bound $(U,V)$ as above since $U_1\cap U_2\subseteq U_2$ and $V_1\cap V_2\subseteq V_2$? – am_11235... Mar 08 '20 at 07:24
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    @am_11235... Yes, the neighbourhoods of a point are directed under reverse inclusion; this is a very commonly used idea in nets. – Henno Brandsma Mar 08 '20 at 07:31