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Taking $B:=\{a \in \Bbb R^{\Bbb N} \mid \exists C \in \Bbb R : |a_n|<C\}$

$J:=\{a \in B \mid \forall n \in \Bbb N : a_{n+1}\geq a_n\}$

and the distance function $d(a,b):=\sup_{n \in \Bbb N}|a_n-b_n|$. I seek to show the set is closed. I have received an answer to this question elsewhere but I wanted to clarify something I don't quite understand about the proof.

Proof:

Consider the compliment of $J, J^C$, then $\exists a_n \in J^C$ s.t. $a_n > a_{n+1}$.

Let $r=a_n-a_{n+1}>0$, then $B_{r/2}(a_n) \subset J^C$. Therefore $J^C$ is open and so $J$ must be closed.

My understanding:

Following all the steps in this proof gives the train of logic that in the compliment there must be a sequence which is bounded but is not non-decreasing as in $J$.

So we can select a sequence which is decreasing instead. Now obviously if the sequence $a_n$ is decreasing then $a_n-a_{n+1}>0$.

Therefore if we take this distance as $r$ and then we take the radius of our open ball to be $r/2$ then the open ball of radius $r/2$ will have as its elements all the sequences which have $a_n>a_{n+1}$ (I think).

My question:

The final line of the proof doesn't really make sense to me, at least not fully. Why do we select $r/2$ instead of just $r$?

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  • $\begingroup$ If $a\in J^c$ it does not follow that $a_n>a_{n+1}$ for some $n$. This inference is only valid for $a \in J^c\cap B$. $\endgroup$ – DanielWainfleet Aug 17 '18 at 0:45
  • $\begingroup$ @DanielWainfleet I had assumed if you specify a metric space and then specify a subspace that the compliment of the subspace implied we were looking at things that are in $J^C$ and $B$ i.e. $J^c\cap B$, is this incorrect ? $\endgroup$ – excalibirr Aug 17 '18 at 15:05
  • $\begingroup$ I think it was unclear to me that you meant that $B$ is the whole space. In which case you are right. I am too used to thinking about $l^{\infty},$ the space of all bounded real sequences $\endgroup$ – DanielWainfleet Aug 17 '18 at 17:02
  • $\begingroup$ @DanielWainfleet I'm glad we are on the same page now at least I had worried I had perhaps not stated it as clearly as may be required I am prone to skipping important statements sometimes :) $\endgroup$ – excalibirr Aug 17 '18 at 19:08
  • $\begingroup$ And I am prone to mis-reading Q's when I stay here too long when I should be asleep. $\endgroup$ – DanielWainfleet Aug 18 '18 at 4:37
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… then $\exists a_n\in J^C$ …

That doesn't make sense. $a_n$ is just an individual real number, so it can NOT be an element of $J^C$, which is a set of sequences. In this context, apparently $a_n$ is a component of such a sequence, if we use the notation $a=(a_n)_{n\in\mathbb{N}}$.

If we want to prove that $J^C$ is open, then we need to show that for any element $a\in J^C$ some ball around $a$ is still contained in $J^C$. So that part instead says the following:

"Consider $J^C$, the compliment of $J$. Pick an arbitrary $a\in J^C$. Then $\exists n\in\mathbb{N}$ s.t. $a_n>a_{n+1}$."

… then $B_{r/2}(a_n)\subset J^C$.

Again, that doesn't make sense because $a_n$ is a number, not a sequence. Just like in the first correction above, this is supposed to be "$B_{r/2}(a)\subset J^C$".

So we can select a sequence which is decreasing instead.

No, we don't. See above — we select an arbitrary sequence in $J^C$, which is the same as saying a sequence that is not in $J$. Such a sequence is not non-decreasing, but that does NOT make it decreasing. Being in $J$ means that $\forall n: a_{n+1}\ge a_n$. The negation of "for all" is "there exists", because even a single violation makes a "for all" property not true. So being in $J^C$ means that $\exists n: a_{n+1}<a_n$, as already stated above.

For example, $a=(1,2,4,3,5,6,6,6,6,6\cdots)\in J^C$, even though it's not decreasing.

Then the open ball of radius $r/2$ will have as its elements all the sequences which have $a_n>a_{n+1}$ (I think).

Well, that's exactly the point, but why is it going to be true?

First of all, to avoid confusion, let's give different names to different sequences: if we started with an original sequence $a\in J^C$, let's say we're looking at an arbitrary sequence $b\in B_{r/2}(a)$, and we want to show that $b\in J^C$ as well. Note that we're not even trying to prove that $b$ is decreasing. But we will achieve our goal if we prove that $b$ is decreasing at least at that one place, i.e. if we prove that $b_n>b_{n+1}$.

By the definition of the norm, we know that $$d(a,b)=\sup_{k\in\mathbb{N}}\left|a_k-b_k\right|<\frac{r}{2}.$$ Therefore, in particular, $$b_n>a_n-\frac{r}{2} \quad \text{and} \quad b_{n+1}<a_{n+1}+\frac{r}{2}.$$ Then $$b_n-b_{n+1}>\left(a_n-\frac{r}{2}\right)-\left(a_{n+1}+\frac{r}{2}\right)=\underbrace{a_n-a_{n+1}}_{r}-r=0,$$ as desired.

Why do we select $r/2$ instead of just $r$?

Because otherwise that last calculation may fail to work.

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  • 1
    $\begingroup$ Thank you this is a great and detailed answer to my question :) I had meant to write $\{a_n\}_{n \in \Bbb N }$ to denote it as a sequence but I carelessly forgot so of course you are right that it does not make sense when I refer to $a_n$ as a sequence when in fact this symbol denotes a real element of the sequence. Thank you for pointing that out ... I need to be more careful about that in future. $\endgroup$ – excalibirr Aug 17 '18 at 0:26
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    $\begingroup$ @exodius: You're welcome! I'm glad I could help. :-) $\endgroup$ – zipirovich Aug 17 '18 at 3:41

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