Show that two topologies coincide on a subset

convergence-divergencegeneral-topologynets

Let $E$ be a set, $\tau_i$ be a topology on $E$ and $B\subseteq E$.

I wonder how we would need to show that $$\left.\tau_1\right|_B:=\{\Omega\cap B:\Omega\in\tau_1\}=\{\Omega\cap B:\Omega\in\tau_2\}=\left.\tau_2\right|_B\tag1.$$

I already know the following:

Lemma 1: The following are equivalent:

  1. $\tau_1$ is coarser than $\tau_2$;
  2. $\operatorname{id}_E$ is $(\tau_2,\tau_1)$-continuous;
  3. If $x\in E$ is a $\tau_2$-limit point of a net $(x_t)_{t\in I}\subseteq E$, then $x$ is a $\tau_1$-limit point of $(x_t)_{t\in I}$.

So, from this we immediately obtain that the following are equivalent:

  1. $\tau_1=\tau_2$;
  2. $\operatorname{id}_E$ is a homeomorphism from $(E,\tau_1)$ onto $(E,\tau_2)$;
  3. If $x\in E$ and $(x_t)_{t\in I}\subseteq E$ is a net, then $x$ is a $\tau_1$-limit point of $(x_t)_{t\in I}$ if and only if $x$ is a $\tau_2$-limit point of $(x_t)_{t\in I}$.

So, I guess we need to find a characterization of the net convergence in a subspace topology. Let $\tau$ be any topology on $E$, $(x_t)_{t\in I}\subseteq B$ be a net and $x\in B$.

Are we able to show that
7. $x$ is a $\left.\tau\right|_B$-limit point of $(x_t)_{t\in I}$;
8. $x$ is a $\tau$-limit point of $(x_t)_{t\in I}$

are equivalent?

This should be correct: Assume (7.) and let $N$ be a $\tau$-neighborhood of $x$. Then there is a $\Omega\in\tau$ with $x\in\Omega\subseteq N$ and, since $x\in B$, $O:=\Omega\cap B$ is a $\left.\tau\right|_B$-neighborhood of $x$. So, by (7.)$, $$\exists t_0\in I:\forall t\ge t_0:x_t\in O\subseteq\Omega\subseteq N;$$ i.e. (8.).

Now assume (8.) and let $N$ be a $\left.\tau\right|_B$-neighborhood of $x$. Then there is a $\Omega\in\tau$ with $x\in O:=\Omega\cap B\subseteq N$ and obviously $\Omega$ is a $\tau$-neighborhood of $x$. So, by (8.), $$\exists t_0\in I:\forall t\ge t_0:x_t\in\Omega.$$ Since $(x_t)_{t\in I}\subseteq B$, we conclude $$\forall t\ge t_0:x_t\in\Omega\cap B=O\subseteq N;$$ i.e. (7.).

So, $(1)$ should be equivalent to

  1. If $(x_t)_{t\in I}\subseteq B$ is a net and $x\in B$, then $x$ is a $\tau_1$-limit point of $(x_t)_{t\in I}$ if and only if $x$ is a $\tau_2$-limit point of $(x_t)_{t\in I}$,

right?

Best Answer

It’s correct, but you can shorten it considerably. For $i\in\{1,2\}$ let $\tau_i^B$ be the relative topology on $B$ induced by $\tau_i$. It’s trivial that if $\tau_1^B=\tau_2^B$, $\nu$ is a net in $B$, and $p\in B$, then $\nu$ converges to $p$ in $\tau_1^B$ iff if $\nu$ converges to $p$ in $\tau_2^B$.

For the converse, suppose that $\tau_1^B\ne\tau_2^B$. Without loss of generality there is some $U\in\tau_1^B\setminus\tau_2^B$. Let $F=B\setminus U$; $F$ is not closed in $\tau_2^B$. Since $F$ is not closed in $\tau_2^B$, there is a $p\in(\operatorname{cl}_{\tau_2^B}F)\setminus F$, and there is a net $\nu$ in $F$ converging to $p$ in $\tau_2^B$. In $\tau_1^B$, however, $U$ is an open nbhd of $p$ that contains no term of the net $\nu$, so $\nu$ does not converge to $p$ in $\tau_1^B$.

Thus, $\tau_1^B=\tau_2^B$ iff $\left\langle B,\tau_1^B\right\rangle$ and $\left\langle B,\tau_2^B\right\rangle$ have the same convergent nets.

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