Why A is closed in M does not imply A has maximum?

Question

After I wrote this question, many people commented that maximum needs to be defined with a metric space and a relevant norm. Let me just simplify my question as only in $\Bbb R^1$:

For a set $A \subseteq\,\Bbb R^1$, in a euclidean metric,
<p>$m = \max(A)$ when $m\in A, \forall x\in A, x \le m$</p>

I know that the following claim is true:

A is compact $\to\;\exists\,\max$(A), and $\sup(A) = \max(A)$

Why is the following claim not true? (The claim is fixed again. I am so sorry)

For A$\subseteq\Bbb R^1$

<p>A is closed in $\Bbb R^1 \;\land\; \exists\,\sup (A) \to\;  \sup$(A) = $\max$(A)</p>

I tried to find already existing questions but could find only one, which does provide a sufficient answer.

Is it not possible for a set to be both closed an unbounded? In $\mathbb{R}$, you might want to consider intervals of the form $[a,\infty)$. Or even just consider the entire space $M$, which is both open and closed (clopen!). — Xander Henderson, Aug 05 '17 at 16:33
I am sorry, I realized I made a mistake in my question. I am fixing the question now. — acubens555, Aug 05 '17 at 16:36
@acubens555 You did not answer my question: what is $\max(A)$ in a metric space? — José Carlos Santos, Aug 05 '17 at 16:48
@acubens555, if you do not tell us what you mean by $\max(A)$ when $A$ is a subset of a metric space, your question is impossible to answer — usually, that does not mean anything. — Mariano Suárez-Álvarez, Aug 05 '17 at 16:53
@JoséCarlosSantos Max(A)=m where ($\exists m\in A, \forall x\in A, x \le m$) — acubens555, Aug 05 '17 at 16:56
@acubens555 Then please tell me what does $x\ge m$ mean in a metric space. — José Carlos Santos, Aug 05 '17 at 16:57
@JoséCarlosSantos x is smaller than m. Is it only possible in $\Bbb R^1$ ? — acubens555, Aug 05 '17 at 16:58
@MarianoSuárez-Álvarez I am sorry for my defective question. I am trying to answer your questions. I have only learned introductory real analysis, so I do not know much. — acubens555, Aug 05 '17 at 17:00
A metric is not enough to define a maximum element. You need something more, i.e. some way of determining the size of an element. That is, you need a norm. In order to say what the "maximal element of $A$" is, you have to define a norm. The metric doesn't do that for you. — Xander Henderson, Aug 05 '17 at 17:05
Your new edit doesn't answer the question---what does it mean for $x \le m$? The complex plane is a metric space, but which is larger: $i$ or $1$? — Xander Henderson, Aug 05 '17 at 17:06
@XanderHenderson Thanks for your comments. I fixed it again. Lets only confine to Real numbers with Euclidean metric. — acubens555, Aug 05 '17 at 17:19
Ever closed and bounded subset of $\mathbb{R}$ has a maximal element, and that maximum is equal to the supremum (indeed, any time a set has a maximal element, that maximal element must also be the supremal element). — Xander Henderson, Aug 05 '17 at 17:20
Under the definitions in your question, what is, for example $\max([1,+\infty))$? — , Aug 05 '17 at 17:21

score 2 · Answer 1 · answered Aug 05 '17 at 17:22

2

For every subset $A$ of $\mathbb R$ (closed or not), if $\max(A)$ exists, then it is equal to $\sup(A)$. In fact, if $m=\max(A)$ then $(\forall a\in A):a\leqslant m$, and therefore $m$ is an upper bound of $A$. Furthermore, if $m'<m$ then there is an element of $A$ which is greater than $m'$, namely $m$. So, $m$ is the least upper bound of $A$, which means that $m=\sup(A)$.

What made you think that this was not true?

answered Aug 05 '17 at 17:22

José Carlos Santos

427,504

Oh... my god.. I did mistake. I am so sorry for many mistakes. Let me fix my question again as below: A is closed in $\Bbb R^1 ;\land; \exists,\sup (A) \to; \exists,\max(A) \land, \max$(A)=$\sup$(A) – acubens555 Aug 05 '17 at 17:27

Xander Henderson · Accepted Answer · 2017-08-05T18:06:14.930

1

Let $A \subseteq \mathbb{R}$ be closed, and suppose that $-\infty < \sup(A) = \alpha < \infty$. By definition of the supremum, for each $n\in\mathbb{N}$, there is some $x_n \in A$ such that $$ \alpha - x_n = |\alpha - x_n| < \frac{1}{n}.$$ But then $x_n$ is a sequence in $\mathbb{R}$ that converges to $\alpha$. This implies that $\alpha$ is a limit point of $A$. As $A$ is closed, it contains all of its limit points, therefore $\alpha \in A$. But then we have $$ \alpha \in A \qquad\text{and}\qquad \alpha \ge x\, \forall x\in A. $$ Therefore $\alpha = \max(A)$.

edited Aug 05 '17 at 18:06

answered Aug 05 '17 at 17:32

Xander Henderson

29,772

I can understand your answer. The existence of sup(A) implies A is bounded above. Since A is closed, sup(A) is also the element of A. So max(A)=sup(A). – acubens555 Aug 05 '17 at 17:39
Essentially, yes. – Xander Henderson Aug 05 '17 at 17:41
Then in what cases A being closed in M is not enough to guarantee sup(A)=max(A) ? I can understand if M = $\Bbb R^1$, than sup(A)=max(A) is guaranteed. – acubens555 Aug 05 '17 at 17:43
The supremum is always defined, but need not be finite. The set $[a,\infty)$ is closed, but not bounded. This set has no maximum, but $\sup[a,\infty) = \infty$ is perfectly well defined. – Xander Henderson Aug 05 '17 at 17:44
Does [a,$\infty$) have supremum? I have never heard about this... Thanks for your teaching. – acubens555 Aug 05 '17 at 17:48
So your example of [a,$\infty$) is an example that the following claim is wrong. I am right? For A$\subseteq\Bbb R^1$, A is closed in $\Bbb R^1 ;\land; \exists,\sup (A) \to; \exists,\max(A) \land, \max$(A)=$\sup$(A) – acubens555 Aug 05 '17 at 17:51
Oh I fixed...Too many obvious mistakes today... – acubens555 Aug 05 '17 at 17:53
Every subset of $\mathbb{R}$ has a supremum. The important point is that the supremum is finite, i.e. the set is bounded. The correct formulation of the theorem is: if $A\subseteq \mathbb{R}$ is closed and bounded (bounded means that $\sup(A) < \infty$), then $A$ has a maximum, and $\max(A) = \sup(A)$. – Xander Henderson Aug 05 '17 at 17:53
Thank you so much!! I can now understand all of my questions. Thanks a lot. – acubens555 Aug 05 '17 at 17:57
I think you need to add A is non-empty, since empty set is closed and bounded, but does not have the supremum. – acubens555 Aug 05 '17 at 18:02
$\sup(\emptyset) = -\infty$, thus the empty set is not bounded. – Xander Henderson Aug 05 '17 at 18:05
A professor in U-Wisconsin Madison at math department confirmed me that empty set does not have the supremum. I am not sure who is right. – acubens555 Aug 05 '17 at 18:07
Let $a\in\mathbb{R}$. It is vacuously true that if $x\in\emptyset$, then $x \le -\infty < a$. Therefore $\sup(\emptyset) < a$. As this holds for all $a\in\mathbb{R}$, it follows that $\sup(\emptyset) = -\infty$. – Xander Henderson Aug 05 '17 at 18:09
Thanks. I will ask your claim to the professor. I am confused. Thanks a lot!! – acubens555 Aug 05 '17 at 18:11

Why A is closed in M does not imply A has maximum?

2 Answers2