Let $x \in X$. If for every net $(x_{\lambda})_{\lambda \in \Lambda}$ in $X$ such that $x_{\lambda} \rightarrow x$ there exists $\lambda_0 \in \Lambda$ such that $x_{\lambda} \in A \ \ \ \forall \lambda \geq \lambda_0$ then $A$ is a neighborhood of $x$.

My solution so far: Suppose that $A \notin \mathcal{N}_x$, where $\mathcal{N}_x$ the family of sets containing $A$, then $x \notin A^o \Leftrightarrow x\in X-A^o=\overline{X-A}$. Since $\overline{X-A}$ Then there exists a net $(x_{\lambda})_{\lambda \in \Lambda}$ in $X$ such that $x_{\lambda} \in \overline{X-A} \ \ \ \ \forall \lambda \in \Lambda$ and $x_{\lambda} \rightarrow x$ which is a contradiction.

I'm not sure if $x_{\lambda} \in cl(X-A)$ indeed is a contradiction.

Another approach is writing down the definition of net convergence and showing that $A \cap U \neq \emptyset$ for every $U \in \mathcal{N}_x$. But this shows (again) that $x\in \overline{A}$, while I would like to show that $x \in A^o$.

Thank you in advance!

  • $\begingroup$ You need to choose $x_{\lambda} \in X - A$, if you take $\overline{X - A}$, you can't guarantee that any $x_{\lambda}$ is outside $A$. $\endgroup$ – Daniel Fischer Jan 2 '18 at 16:30

If $A$ is not a neighbourhood of $x$, this means that for any open set $O$ that contains $x$, $O \nsubseteq A$. So assume for a contradiction that this is the case. A direct proof is quite doable:

Now let $\Lambda$ be the set of all open sets containing $x$ in reverse inclusion ordering, which makes it into a directed set.

Now for each $O \in \Lambda$ pick $x_O \in O\setminus A$, by the non-inclusion we assumed.

If $O$ is open containing $x$, $O \in \lambda$, and for all $\lambda=O' \ge O$, $x_\lambda \in O' \subseteq O$, so for all $\lambda \ge O $ we have $x_\lambda \in O$. So $x_\lambda \to x$. But by construction, $x_\lambda \notin A$ for all $\lambda$. This contradicts the given property of $A$. So $A$ must be a neighbourhood of $x$ after all.

In your proof you can pick all $x_\lambda \in X-A$ straight away, because of the fact that $x \in \overline{B}$ iff there is a net from $B$ that converges to $x$, as a general fact (part of which I reproved above, in fact).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.