Open neighbourhoods & polish spaces - typo in Marker's notes?

Question

A very (very!) easy question that merges together the very basic concepts of two fields that I find always problematic for my understanding, namely topology and descriptive set theory. Everything starts from a passage taken from the – rightly praised – notes written by David Marker on descriptive set theory.

If $\sigma \in \mathbb{N}^{<\omega}$, let $N_{\sigma} := \{ f \in \mathcal{N} \ : \ \sigma \subset f \}$. Then $N_\sigma$ is an open neighborhood of $f$.

In general I find that being able to spot a typo is a good measurement of the understanding of a certain field or topic. In those topics I am completely at loss, and I do not trust my logic, but indeed I think this is a typo (most probably one of those typos that a reasonable reader could spot and go on without much thoughts). Considering I do have doubts, here there is my line of reasoning...

The correct statement should be that $N_\sigma$ is an open neighbourhood of $\sigma$, because we endow $\mathcal{N}$ with the discrete topology and thus every $f \in \mathcal{N}$ is open. Given a topological space $(X, \tau)$, a neighbourhood $N_x$ of a point $x$ is deemed open when there is an open set $G \in \tau$ such that $x \in G \subset N_x$. Thus in our case, every $f$ is an open set that contains $\sigma$ and that is also a subset of $N_\sigma$, where the union of all those open sets given by the various $f$ is $N_\sigma$ itself.

Questions

• Am I right? Is the logic sound?

• If this is not the case, why is $N_\sigma$ an open neighbourhood of $f$?

As always, any feedback is most welcome.
Thank you for your time.

Answer 1

No, $\sigma$ is not even an element of the space $\mathcal N$. We do not endow $\mathcal N$ with the discrete topology. Notation $\sigma \subset f$ means $\sigma$ is a restriction of $f$. So this open set $N_\sigma$ consists of all infinite $f$ that extend the finite recipe $\sigma$.

For example, take this case: $\sigma := \{(0,17)\} \in \mathbb N^1$. Then $N_\sigma$ is all $f \in \mathbb N^\omega$ such that $f(0)=17$. This is an open set for the product topology of $\mathcal N = \mathbb N ^ \omega$.

Answer 2

GEdgar's answer is being a little too kind: there clearly is a mistake in the notes. The correct statement is just that ${\cal N}_{\sigma}$ is open in $\cal N$. I think that the author may have meant to write something like "${\cal N}_{\sigma}$ is an open neighbourhood of any if its elements $f$". In your (accurate) quotation the second $f$ is an unspecified free variable and that makes the statement obviously wrong.