Show that the set of all bounded sequences in $X=\mathbb{R}^\Bbb N$ with the box topology is an open and closed set.

Question

Show that the set of all bounded sequences in $X=\mathbb{R}^\Bbb N$ with the box topology is an open and closed set.

I thought of two ways of how to approach this. First one was to note that this set is $$S=\{(x_n)_{n \in \Bbb N} : |x_n| < M, \forall n \in \Bbb N\}$$ for some $M$ a natural number. Now since $|x_n| < M \iff - M < x_n < M$ I can say that $$S=\{(x_n)_{n \in \Bbb N} : -M< x_n \} \cup \{(x_n)_{n \in \Bbb N} : x_n < M\}$$

And now these sets are the preimages (for all $n$) $$\pi_n^{-1}(-M, \infty) \cup \pi_n^{-1}(-\infty, M)$$ which are both open I think?

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Another approach was to and show that if $(x_n)_{n \in \Bbb N} \in A$, then finding an open set $U$ in $A$ such that for any $y_n \in U$ we have that $y_n \in A$.

I think that this one would be the correct approach as the former does not use that fact that we're assuming the box topology.

A basic open set (w.r.t box topology) in $A$ is of the form $$\prod_{n \in \Bbb N}U_n$$ with $U_n$ being open in $A_n$ (assuming $A = \prod_{n \in \Bbb N} A_n)$. And $U_n$ is open in $A_n$ if and only if $U_n = V_n \cap A_n$ for $V_n$ open in $X_n$.

So the basic open set is of form $$\prod_{n \in \Bbb N}(O_n \cap A_n)$$

How can I find such a set which satisfies the wanted property?

Answer 1

Your first approach doesn't really work. The first problem is that the set $S$ you define is not the set of bounded sequences. Rather, it is the set of sequences bounded by $M$. A second problem occurs in your second equality, where the $\cup$ should be a $\cap$. Finally, you find $$\bigcap_{n\in\mathbb{N}}\left(\pi_{n}^{-1}(-M,\infty)\cap \pi_{n}^{-1}(\infty,M)\right)$$ but this need not a priori be open, as it is an infinite intersection of open sets.

A good approach would be to show that a sequence $(x_{n})_{n\in\mathbb{N}}$ is bounded if and only if sequences that are 'close by' are also bounded. For example, let $U_{i}$ be a small open ball around $x_{i}$. Then $U=\prod U_{i}$ is an open ball around $(x_{n})_{n\in\mathbb{N}}$. Can you show that any sequence in $U$ is bounded?

Answer 2

In the first approach $$S =\cup_M \cap_n\pi_n^{-1}(-M, \infty) \cap \pi_n^{-1}(-\infty, M)$$ and intersction of a sequence of open sets is not open in general.

Hint for second approach: If $(x_n)$ is a bounded sequence and $(y_n) \in U_1\times U_2\times\dots $ where $U_n=(x_n-1,x_n+1)$ then $|y_n-x_n| <1$ for al $n$ , so $(y_n)$ is also a bounded sequence.

If, on the other hand, $(x_n)$ is unbounded then $(y_n)$ is also unbounded. This proves that the set of all bounded sequences and the unbounded sequences are both open.

Answer 3

Both approaches work to show that the set of bounded sequences is open. One solves it by describing the set as a union of open sets, but I think you should formulate it a little differently. Specifically, we have for each positive (real or natural) number $M$ the open set $$ S_M=\{(x_n)\mid \forall n\in\Bbb N:|x_n|<M\} $$ and the set of all bounded sequences is the union of these.

The other approach works by explicitly finding a neighborhood around each sequence. We can finish your neighborhood approach the following way: Let $(x_n)$ be a bounded sequence, and let $$U=\{(y_n)\in X\mid \forall n\in\Bbb N:|x_n-y_n|<1\}.$$This is an open neighborhood (it is easy to see that this $U$ is a box, i.e. a basic open), and any sequence in $U$ is obviously bounded.

Showing that the complement is open is, in my opinion, most easily shown analogously to the second approach: for an unbounded sequence $(x_n)$, the neighborhood $U$ defined in the exact same way necessarily contains only unbounded sequences.