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My question reads

If $A$ and $B$ are nonempty, bounded, and satisfy $A\subseteq B$, then $\sup A\le \sup B$.

To me this makes sense because $A$ is a subset of $B$, but I am having issues setting up my proof.

Is it correct to say that because $A$ and $B$ are bounded that they each have a supremum and then use this fact? I was thinking of saying let $a\in A$ be the supremum of $A$ and $b\in B$ be the supremum in $B$. Since $A$ is a subset of $B$, it follows that $a\in\ B$ as well and since $b$ is the supremum of $B$ we have that $a\leq\ b$. Hence, $\sup A\leq \sup B$.

Updated Proof:

Suppose $\operatorname{sup}A>\operatorname{sup}B$. Then, $\operatorname{sup}B$ is not an upper bound for A. Then there exists an $x\in\ A$ such that $x>\operatorname{sup}B$. By subset, $x\in\ B$ where $x>\operatorname{sup}B$, which tells us that $\operatorname{sup}B$ is not an upper bound for $B$. This is a contradiction.

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    $\begingroup$ Your intuition is exactly right because the supremum acts like a maximum. Unfortunately it isn't really a maximum because it may lurk on the edge of the set without actually belonging to it. Still, it is quite helpful to think of a supremum as a "generalized maximum" for a notion of the direction for your proof, then adjust your proof as needed. $\endgroup$ – MPW Sep 14 '17 at 2:30
  • $\begingroup$ discrete-mathematics is a wrong tag for your question. One should definitely not use it here. $\endgroup$ – Jack Sep 18 '17 at 17:05
  • $\begingroup$ Try to respect the notation always, even in digital format. That is, $sup\rightarrow \operatorname{sup}$ (or some other format). The phrase "By subset" doesn't tell anything; of course we understand it, but the point is that everything has to be complete and explicitly said. $\endgroup$ – user2820579 Mar 24 at 17:13
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proof-verification

Is it correct to say that because $A$ and $B$ are bounded that they each have a supremum and then use this fact? (Well, let's see how you use it.)

I was thinking of saying let $a\in A$ be the supremum of $A$ and $b\in B$ be the supremum in $B$. (No. The supremum of the set A does not need to belong to A. Consider for instance A = (0,1). Your argument breaks down here.) Since $A$ is a subset of $B$, it follows that $a\in\ B$ as well and since $b$ is the supremum of $B$ we have that $a\leq\ b$. Hence, $\sup A\leq \sup B$.


To give a correct proof, try the following "easier" exercises:

  1. Show that both $\sup A$ and $\sup B$ exist. (Well, you have known how to do this.)
  2. Show that for any $b\in B$, we have $b\leq\sup B$. (Please pay close attention to how I use the letter $b$ differently: it just means an element of the set B, not the supremum of it. One could equivalently say "for any $x\in B$, $x\leq \sup B$" or "for any $\epsilon\in B$, $\epsilon\leq \sup B$".)
  3. Using (2) and the assumption $A\subset B$ to show that for any $a\in A$, we have $a\leq \sup B$.
  4. Using (3) to finish the proof: $\sup A\leq \sup B$.

[Added later:]

Let $A$ and $B$ be nonempty, bounded sets such that $A\subseteq B$. Now suppose instead that $\sup A>\sup B$. Then, $\sup B$ is not an upper bound for $A$. (True. But why? Well, because sup A is the "least" upper bound for A.) Then there exists an $x\in\ A$ such that $x>\sup B$. By subestSince A is a subset of B, we have that $x\in\ B$ where $x>\sup B$, which tells us that $\sup B$ is not an upper bound for $B$, and hence not a least upper bound for $B$. This is a contradiction.

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  • $\begingroup$ Thank you for the suggestions! I actually think I get it now and used contradiction instead. I will edit my proof and hopefully it will make sense now. $\endgroup$ – Sam Sep 14 '17 at 21:19
  • $\begingroup$ @Sam You are welcome! Since you have put the proof-verification tag to your post, you might want to leave the question as it was so that it would not invalidate any "proof verification" answers. You could add further editing under you wrote. $\endgroup$ – Jack Sep 14 '17 at 21:36
  • $\begingroup$ Does what I have now make sense? $\endgroup$ – Sam Sep 15 '17 at 0:32
  • $\begingroup$ @Sam: one step needs further explanation. That's easy to fix. $\endgroup$ – Jack Sep 15 '17 at 1:06
  • $\begingroup$ Okay, so I need to mention that fact of sup A is least upper bound for A? $\endgroup$ – Sam Sep 15 '17 at 1:54
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A very direct proof:

Lemma 1: If $w$ is an upper bound of $X$ then $w \ge \sup X$.

Pf: This is almost the definition. It is the contrapositive of definition. If $w < \sup X$ it can't be an upper bound.

Lemma 2: If $A \subset B$ and $w$ is an upper bound of $B$, then $w$ is an upper bound of $A$.

Pf: Immediate consequence of the definitions. For all $a \in A$ then $a \in B$ and $w \ge a$ because it is an upper bound of $B$. So it is an upper bound of $A$.

Now the statement is obvious.

$w = \sup B$ is an upper bound of $B$ $\implies w$ is an upper bound of $A \subset B$ (Lemma 2) $\implies \sup B = w \ge \sup A$ (Lemma 1).

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It's not true that a bounded subset of $\mathbb R$ contains its supremum: take the bounded interval $I = (0,1)$ and note that $\sup I = 1 \notin I$.

The definition of supremum of a bounded set is: the least upper bound of the set, that is, if $A\subseteq\mathbb R$ is bounded, then both $$\sup A \geq a \; \; \forall a\in A$$ and $$ x\geq a \; \; \forall a\in A \;\; \Longrightarrow \;\; \sup A \leq x.$$ Now, note that if $\sup B \geq b$ for every $b\in B$, then (since $A\subseteq B$) $\sup B\geq a$ for every $a\in A$. Therefore $\sup A \leq \sup B$.

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Since $b\leq\sup B$ for all $b\in B$, then of course $a\leq\sup B$ for all $a\in A$ (since elements of $A$ are also elements of $B$).

In other words, $\sup B$ is an upper bound for $A$. Therefore $\sup A$, which is the least upper bound for $A$, is no larger.

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  • $\begingroup$ If $K$ is any upper bound for $B$ (whether or not it's the least one), then $b\leq K$ for all $b\in B$. That's what it means to be an upper bound, right? As I said in my comment above, think of $\sup B$ as if it were $\max B$. You probably don't have any difficulty accepting the statement that $b\leq \max B$ for all $b\in B$, do you? Same thing, just on the "upper edge" of the set. $\endgroup$ – MPW Sep 14 '17 at 2:36
  • $\begingroup$ Think of putting a lasso loosely around the set, maybe with a gap above. That gives an upper bound. Then tighten it a little. You get a smaller upper bound, maybe with a smaller gap above. Continue. When the lasso is pulled as tight as possible, the upper bound has shrunk as low as it can, still surrounding the set, and you have reached the supremum. Ditto for the infimum, if you look at the lower end of the set as you tighten it up, but it will rise as high as it can from the bottom. $\endgroup$ – MPW Sep 14 '17 at 2:46
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I was thinking of saying let $a\in\ A$ be the suprema of A and $b\in\ B$ be the suprema in B. Since A is a subset of B, it follows that $a\in\ B$ as well and since $b$ is the suprema of $B$ we have that $a\leq\ b$. Hence, $supA\leq\ supB$.

No, the supremum of $A$ is not necessarily an element of $A$, and so need not be an element of $B$.   The supremum is the least value that is greater or equal to any element of the set.

$$a=\sup A ~\iff~ \forall x\in A~( x\leq a )~\wedge~ \neg\exists y\notin A~\forall x\in A : (x\leq y < a)$$

$$b=\sup B ~\iff~ \forall x\in B~( x\leq b )~\wedge~ \neg\exists y\notin B~\forall x\in B : (x\leq y < b)$$

So attack from the other set.

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