Let $X= \Bbb R^{\Bbb N}$ and let $A\subset X$ be the set of bounded sequences. Is $A$ open/closed in the product topology of $\Bbb R^{\Bbb N}$?

Question

Let $X= \Bbb R^{\Bbb N}$ and let $A\subset X$ be the set of bounded sequences. Is $A$ open/closed in the product topology of $\Bbb R^{\Bbb N}$?

I have hard time with these questions related to the product topology. It seems that there is no simple answer to what the open sets of the product topology are. I know that the basis is formed by $$B= \bigcap_{j \in K} \pi_j^{-1}(O_j)$$ where $K \subset J$ is a finite indexing set. So by definition the open sets of the product topology are unions of these intersections, but I still don't have an explicit form for an open set.

For example in this question how I would naturally approach it is to pick $a \in A$ and then consider some neighborhood $O_a$ and determine if it's contained in $A$. If so, then I can conclude that $A$ is open, but I don't have any form for $O_a$ so I cannot really work with it. What I could work with is that if $O_a$ is a neighborhood of $a$, then by definition there exists a basis element $B$ such that $$a \in B \subset O_a$$ So know I have that $a \in \bigcap_{j \in K} \pi_j^{-1}(U_j)$ for some $U_j$ open in $\Bbb R$.

I also know that the basis elements of the product topology seem to have some property that $B_j=X_j$ for all, but finitely many $j$, but this I don't really understand. What can I do with these kind of problems?

Answer 1

Convergence in the product topology is just convergence of each coordinate.

Consider $\{1,0,0...\},\{1,2,0,0,...\}, \{1,2,3,0,0,...\},...$. This gives a sequence of elements in the space each of which belongs to the subspace of bounded sequences. This sequence converges in the product topology to $\{1,2,3...\}$ which is unbounded. Hence, the subspace is not closed. On then other hand, $\{0,1,2,3...\},\{0,0,1,2,3...\}, \{0,0,0,1,2,3,...\},...$ gives a sequence in the complement which converges to the bounded sequence $\{0,0,...\}$. So the complement is not closed either. Hence, the given subset is neither closed not open.

Answer 2

Suppose that $O=\bigcap_{n \in K}\pi_n^{-1}[O_n]$ is any non-empty basic open set of $\Bbb R^{\Bbb N }$, so $K \subseteq \Bbb N$ is finite and $O_n$ is some non-empty open subset of $\Bbb R$. Pick some $p_n \in O_n $ arbitrarily. Note that to be in $O$ we just have finitely many conditions on the sequence, i.e. the $n$-th coordinate should be in $O_n$ for all $n \in K$. We can exploit that:

Define $q_n = p_n, n \in K; q_n=0, n \notin K$ and this defines a point $q \in A \cap O$ (the sequence is obviously bounded as it only has at most finitely many non-zero components and the coordinates in $K$ are indeed in $O_n$).

Similarly, define $r_n = p_n, n \in K; r_n = n, n \notin K$. Then this is a point $r \in A^\complement \cap O$ by similar reasoning.

So any non-empty basic open subset and hence any non-empty set, intersects both $A$ and its complement. So $A$ is not closed and not open, $A$ and its complement are even both dense in $\Bbb R^{\Bbb N}$.

This is in contrast with the box topology where $A$ is clopen (closed-and-open).