Let $X$ be a set and $\mathcal{T_1,T_2}$ two Hausdorff topologies on $X$ such that they admit the same convergent nets, i.e., a net $(x_{\alpha})_{\alpha}$ in $X$ converges with respect to $\mathcal{T_1}$ iff it converges with respect to $\mathcal{T_2}$ (but to a possibly different point). Does it follow that $\mathcal{T_1=T_2}$?

If we require that convergent nets converge to the same point, then the result follows easily. If we don't require Hausdorff-ness, then the result is simply false, as the Sierpiński topology and the trivial topology on the set $\{a,b\}$ shows (where every net converges to some point). However, I wonder whether it's true for Hausdorff spaces?

  • 1
    $\begingroup$ A convergent sequence $x_1,x_2,x_3,\dots$ converges to the point $x$ if and only if the sequence $x_1,x,x_2,x,x_3,x,\cdots$ is convergent. See if you can to something similar with nets. $\endgroup$ – GEdgar Jan 31 at 13:22

Let $(x_{\alpha})_{\alpha\in A}$ be a net converging in to $x$ in $\mathcal{T}_{1}$ and to $y$ in $\mathcal{T}_{2}$. We define $$\hat{A}=A\times\{1,2\}$$ and we say $(\alpha,i)\leq(\beta,j)$ iff $\alpha\lneq\beta$ or $\alpha=\beta$ and $i\leq j$. Note that $\hat{A}$, $A\times\{1\}$ and $A\times\{2\}$ are directed sets. We define the net $(y_{(\alpha,i)})_{\hat{A}}$ by $$y_{(\alpha,i)}=\begin{cases}x_{\alpha}&\text{ if }i=1\\ y&\text{ if }i=2\end{cases}.$$

Clearly $(y_{(\alpha,i)})_{\hat{A}}$ converges to $y$ in $\mathcal{T}_{2}$, but in $\mathcal{T}_{1}$ the subnet $(y_{(\alpha,i)})_{A\times\{1\}}$ converges to $x$ and the subnet $(y_{(\alpha,i)})_{A\times\{2\}}$ converges to $y$.

Thus a net converges in $\mathcal{T}_{1}$ if and only if it converges in $\mathcal{T}_{2}$ convergent nets converge to the same point.

  • $\begingroup$ $A\times\{1\}$ is cofinal in $\hat{A}$. Let $(\alpha,i)\in\hat{A}$, there exists a $\beta\in A$ such that $\alpha\leq\beta$ as $A$ is a directed set. Then $(\beta,1)\geq(\alpha,i)$. $\endgroup$ – Floris Claassens Jan 31 at 13:38
  • $\begingroup$ So $(\alpha,1)\leq(\alpha,2)$ and $(\alpha,1)\geq(\alpha,2)$? In that case $(y_{(\alpha,i)})_{\hat{A}}$ may not converge to $y$... $\endgroup$ – Colescu Jan 31 at 13:42
  • $\begingroup$ Good point, no it does not, I was a bit sloppy in writing that comment. You need to take a $\beta\in A$ such that $\alpha\lneq\beta$. Note that if such a $\beta$ does not exists, then $\alpha$ is an upper bound for $A$ in which case $x_{\alpha}=x=y$. $\endgroup$ – Floris Claassens Jan 31 at 13:52
  • $\begingroup$ I see, so it works! Thanks for the elegant solutoin. :) $\endgroup$ – Colescu Jan 31 at 13:56

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