# Two Hausdorff topologies on the same set with the same convergent nets

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?

• 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. – 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.
• $A\times\{1\}$ 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 $(\beta,1)\geq(\alpha,i)$. – Floris Claassens Jan 31 at 13:38
• So $(\alpha,1)\leq(\alpha,2)$ and $(\alpha,1)\geq(\alpha,2)$? In that case $(y_{(\alpha,i)})_{\hat{A}}$ may not converge to $y$... – Colescu Jan 31 at 13:42
• 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 $x_{\alpha}=x=y$. – Floris Claassens Jan 31 at 13:52