trying to prove this set is open/closed

Question

on the set $B=\{a \in \Bbb R ^\Bbb N \mid \exists C \in \Bbb R \forall n \in \Bbb N :|a(n)|<C\}$ we consider the distance $d(a,b):= \sup_{n \in \Bbb N}|a(n)-b(n)| $

for $a,b \in B.$ Let $J=\bigl\{a \in B \mid \forall n \in \Bbb N:a(n+1)\geq a(n) \bigr\}$ be the set of bounded increasing sequences of real numbers.

I want to figure out whether J is open or closed in $(B,d)$.

I thought to do this by constructing a continuous function $f:B \rightarrow Y$ s.t. $Z\subset Y$ and $f^{-1}(Z)=J$. and using the openness/closedness of Z to prove the openness/closedness of J.

So would this be correct ... let $Y=\Bbb R$ then let $Z=(-\infty,C)$ where $C\in \Bbb R$. then take $f: a_n \mapsto c$ where c is the maximum value obtained by the sequence. So as Z is open then so is $f^{-1}(Z)=J$.

Another method I thought to use was as J is a subset of $\Bbb R^{\Bbb N}$ and a subset of $\Bbb R^{\Bbb N}$ is closed iff whenever $a_n$ is a sequence of elements of J and $a_n\rightarrow a$ then a is also an element of J to say that as J is increasing and bounded it is converging, but it does not include its bound as $|a(n)|<C $ so a is not in $J$ and so $J$ is open.

Do either of these look on the right track ?

Answer 1

I think that it's much easier to use the definition:

1. The set $J$ is not open because $0\in J$ (where $0$ is the null sequence), but no open ball centered at $0$ is contained in $J$. In fact, if $r>0$, the ball $B_r(0)$ contains the sequence $\left(\frac r2,\frac r4,\frac r2,\frac r4,\frac r2,\frac r4,\ldots\right)$
2. The set $J$ is closed because if $(a_n)_{n\in\mathbb N}\not\in J$, then there is a natural $n$ such that $a_{n+1}<a_n$. If $r=a_n-a_{n+1}$, then the ball $B\left((a_n)_{n\in\mathbb N},\frac r2\right)$ is contained in $B^\complement$. In fact, if $(b_n)_{n\in\mathbb N}\in J$, then $b_{n+1}<b_n$.

Answer 2

Let me point at the flaws of your argumentation.

For the first part, your map $f\colon B \to \mathbb{R}, (a_n) \mapsto c$ is not wellldefined, since it's possible that a sequence has no maximal value. A well defined map would be $(a_n) \mapsto \sup a_n$.

Then $f^{-1}(Z)$ is open, but it is not $J$. $f^{-1}(Z)$ contains all sequences which are bounded from above by $C$. This is neither a sub- nor a supset of $J$.

On your second argument, for $J$ to be closed, you could show that for a sequence $(a^k)$ inside $J$ which converges to an $a\in B$, the limit $a$ already lies in $J$. Caution: each $a^k$ is itself a sequence. We are dealing with sequences of sequences. So let's assume we have $\lim_{k\to \infty}a^k=a$, this is by definition $\lim_{k\to \infty}a^k_n=a_n$ for all $n\in \mathbb{N}$. Since $a_n^k \leq a_{n+1}^k$ for all $k\in \mathbb{N}$, we obtain $a_n \leq a_{n+1}$ for all $n$. So $a$ lies in $J$, so $J$ is closed. This has a priori nothing to do with being open.