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.


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 $r$?

  • $\begingroup$ 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

… 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.

  • 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
  • 1
    $\begingroup$ @exodius: You're welcome! I'm glad I could help. :-) $\endgroup$ – zipirovich Aug 17 '18 at 3:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.