Let $X=\{0,1\}^{\mathbb{N}}$ be equipped with the product topology. I wonder whether for a fixed $n\in\mathbb{N}$, the subset $Y=\{A\subseteq \mathbb{N}: |A|=n\}\subseteq X$ is closed? I convinced myself that $Y$ is not open. $Y$ consists of sequences of the form $\{ (a_i)_{i=1}^{\infty}: \exists i_1,\ldots,i_n \text{ s.t. } a_{i_1}=\ldots=a_{i_n}=1 \text{ and } a_i=0 \text{ for all } i\notin \{i_1,\ldots, i_n\} \} $.
I would be happy for any help!
On
This set is almost closed: when $x=(x_n)_n \notin Y$ maybe $x$ represents a set with at least $n+1$ elements, and we can fix those in a basic open set, so that any set in it has also at least $n+1$ many elements.
The problem is with sets with fewer elements: any basic neighbourhood around them will contain an $n$-element set because we have "too much freedom". So for $n \ge 1$, $Y_n$ is not closed but $\bigcup_{k=1}^n Y_k = \{A \subseteq \Bbb N: |A| \le n\}$ is closed.
Neither set is open, clearly.
$Y$ is closed when $n = 0$, since it is a singleton. But for $n>0$, $Y$ is not closed.
For example, when $n = 1$, consider the sequence $(\{k\})_{k\in \mathbb{N}}$. This is a sequence from $Y$ which converges to $\varnothing\notin Y$.