Closed set in Baire space

346 Views Asked by At

I am reading a book on Banach spaces. It introduces the Baire space $\mathcal{N}=\mathbb{N}^\mathbb{N}$ as the product of infinitely many copies of $\mathbb{N}$ with the discrete topology.

We have $\mathcal{N}$ is Polish.

In the proof of a result it uses the following: If we let $[\mathbb{N}]$ be the set of infinite subsets of $\mathbb{N}$, then this is a closed subspace of $\mathcal{N}$. Could anyone please explain why this is true?

1

There are 1 best solutions below

8
On BEST ANSWER

"Naturally identified" leaves enough wiggle room to identify an infinite subset $X$ of $\mathbb N$ with the sequence in $\mathcal N$ that enumerates the elements of $X$ in increasing order. The set of all these increasing sequnces is a closed subset of $\mathcal N$.