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Having trouble to prove the following statements regarding topological spaces:

Let $(X,\mathcal T)$ be a topological space.
a) Show that for all mutually distinct points $x,y \in X$ there are open sets $U,V$ with $U\cap V = \emptyset$ and $x\in U, y \in V$ if, and only if, no net in $X$ converges to two different values.
b) Let $\mathcal T'$ be a topology on $X$ and for all nets in $X$ it holds that they converge in $(X, \mathcal T)$ if, and only if, they converge in $(X, \mathcal T')$. Show that $\mathcal T = \mathcal T'$.

Anyone can help?

Martin Sleziak
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lasik43
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  • The one direction in a) should be clear. For the other direction, assume the space is not Hausdorff. By definition there are …. Use that to construct a net converging to two distinct points. – Daniel Fischer May 30 '16 at 11:37
  • Thanks, I'll try that. Can you elaborate a bit on "there are..."? And what about b) ? – lasik43 May 30 '16 at 14:20
  • Take the definition of a space being Hausdorff, and negate it. You get "there are $x,y \in X$, with $x\neq y$ and …". About b): the equivalent assertion using filters instead of nets is rather immediate (because there is a coarsest filter converging to $x$). Try to translate that filter into a net. – Daniel Fischer May 30 '16 at 14:25
  • In the lecture I am attending we didn't cover the notion of filters, so I doubt I will be able to prove the proposition using filters apart from the fact that I am probably not allowed to use them. – lasik43 May 30 '16 at 14:28
  • After the translation, filters wouldn't be mentioned any more. But thinking about it, for nets the most straightforward way is a different one than for filters. Suppose $\mathcal{T}\neq \mathcal{T}'$. Then there is a point $x\in X$ such that the neighbourhood filters of $x$ with respect to $\mathcal{T}$ and $\mathcal{T}'$ differ. Without loss of generality, there is a neighbourhood $V$ of $x$ with respect to $\mathcal{T}$ that isn't a neighbourhood of $x$ with respect to $\mathcal{T}'$. Use $V$ to find a net that converges to $x$ in one topology, but not the other. – Daniel Fischer May 30 '16 at 14:40
  • In your last sentence you mean net, not filter, right? Thanks for your help so far! – lasik43 May 30 '16 at 14:44
  • Right. (Fixed now) – Daniel Fischer May 30 '16 at 14:45

2 Answers2

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Suppose that all distinct $x$ and $y$ have disjoint neighbourhoods, and suppose that $(x_i)_{i \in I}$ is a net converging to both $x$ and $y$. By assumption on $X$ find open $U$ with $x \in U$ and open $V$ with $y \in V$ such that $U \cap V = \emptyset$.

Now apply the definition of convergence to $x$ and $U$. So there exists $i_0 \in I$ such that ...

Same for $y$ and $V$, getting a $j_0 \in I$ such that ...

Now (as $I$ is directed!) find $k \in I$ with $k \ge i_0, k \ge j_0$. Where must $x_k$ be?

For the other direction, suppose every convergent net in $X$ has a unique limit. Suppose now that the disjoint neighbourhood property fails for $X$, so there are $x \neq y$ in $X$ such that for all neighbourhoods $U$ of $x$ and all neighbourhoods $V$ of $y$ we have $U \cap V \neq \emptyset$.

Now define a net: $I = \{(U,V): U \in \mathcal{N}_x, V \in \mathcal{N}_y \}$, ordered by reverse inclusion in both coordinates, and let $f((U,V))$ be any point in $U \cap V \neq \emptyset$ (yes, we use the axiom of choice!). This $f$ defines a net in $X$ (i.e. a map from $I$, the directed set, to $X$; $f(i)$ is denoted $x_i$ as well if the context is clear, as I did above).

Now check (!) that this net converges to both $x$ and $y$ and we have our contradiction.

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If two topologies $\mathcal{T}_1,\mathcal{T}_2$ on $X$ are different, there is at least one subset that is open in one, say $O \in \mathcal{T}_1$, but not in the other, so $O \notin \mathcal{T}_2$. So there is some $x \in O$ that is not an interior point w.r.t. $\mathcal{T_2}$. This means that for every neighbourhood $N$ of $x$ (in $\mathcal{T}_2$), we have $N \nsubseteq O$, so there are always points in $N \setminus O$.

Now there is a standard filter that can now be constructed that converges to $x$ in $\mathcal{T}_2$ and does not converge to $x$ under $\mathcal{T}_1$...

Henno Brandsma
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It is a known result that $(X, \mathcal{T})$ is Hausdorff if and only if the diagonal $\Delta = \{(x,x) : x \in X\}$ is a closed set in $X \times X$ with respect to the product topology.

Let $((x_\lambda, x_\lambda))_{\lambda\in\Lambda}$ be a net in $\Delta$ which converges to $(x,y) \in X \times X$. Then, by the property of the product topology, $(x_\lambda)_{\lambda \in \Lambda}$ converges to both $x$ and $y$ in $X$.

If $X$ is Hausdorff, then $\Delta$ is closed, so $(x,y) \in \Delta$ which implies $x = y$, therefore the limit is unique. Conversely, if nets in $X$ have unique limits, then necessarily $x = y$ so $(x,y) \in \Delta$, implying that $\Delta$ is closed in $X \times X$.

Assume that $x_\lambda \xrightarrow{\mathcal{T}_1} x$ implies $x_\lambda \xrightarrow{\mathcal{T}_2} x$.

We have:

\begin{align} A \text{ is closed in } \mathcal{T}_2 &\iff A \text{ contains all limits of its convergent nets with respect to }\mathcal{T}_2\\ &\implies A \text{ contains all limits of its convergent nets with respect to }\mathcal{T}_1\\ &\iff A \text{ is closed in } \mathcal{T}_1 \end{align}

Therefore, all closed sets in $\mathcal{T}_2$ are also closed in $\mathcal{T}_1$. By taking complements, we conclude $\mathcal{T}_2 \subseteq \mathcal{T}_1$.

mechanodroid
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