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Let $S \subset \mathbb{R}$ (real numbers) be a bounded subset and let $A \subset \mathbb{R}$ be a non-empty subset of $S$. Prove that $\inf S \le \inf A \le \sup A \le \sup S$.

Having trouble understanding this. I get that obviously the $\inf S$ and $\inf A$ will be less than $\sup A$ and $\sup S$, but I don't understand why $S$ and $A$ are different. If I could visualize $S$ and $A$ on a number line that would help a lot. And then to prove this I can't right now as I need more understanding.

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    $\begingroup$ Some simple examples to guide your thinking: $S = [1,2] \subset [0,3] = A$, $S = (0,1) \subset [0,1] = A$ and $S = (0,1) \cup (2,3) \subset [0,9] = A$. $\endgroup$ – BaronVT Sep 10 '19 at 0:50
  • $\begingroup$ The thing is if $k$ is a lowerbound of $S$ it must be a lower bound of $A$ because $A$ is subset of $A$ and if $k$ is less than every element of $S$ then it is lower than every element of $A$ because every element of $A$ is an element of $S$.... The reverse isn't true because if $m$ is a lower bound of $A$ and $m$ is lower than every element of $A$, there could be elements in $S$ that are not in $A$ that are much lower.... An example could but $A = (2,3) \subset (1,4)= S$. So $\inf S=1 < \inf A=2< \sup A =3 < \sup S = 4$. $\endgroup$ – fleablood Sep 10 '19 at 1:11
  • $\begingroup$ I am confused, how is $A = (2,3) \subset (1,4) if neither of the elements in A are in S? $\endgroup$ – Richard Smith Sep 10 '19 at 1:16
  • $\begingroup$ $A$ is the open interval $(2,3)$. It is all the numbers between $2$ and $3$. $S$ is the open interval $(1,4)$. It is all the numbers between $1$ and $4$. If $x \in A$ then $2 < x < 3$ and so $1 < 2 < x < 3 < 4$ so $1 < x < 4$. So $x \in S$. Thus $A\subset S$. $\endgroup$ – fleablood Sep 10 '19 at 1:26
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The thing is if $k$ is a lowerbound of $S$ it must be a lower bound of $A$ because $A$ is subset of $A$ and if $k$ is less than every element of $S$ then it is lower than every element of $A$ because every element of $A$ is an element of $S$.... The reverse isn't true because if $m$ is a lower bound of $A$ and $m$ is lower than every element of $A$, there could be elements in $S$ that are not in $A$ that are much lower.... An example could but $A = (2,3) \subset (1,4)= S$. So $\inf S=1 < \inf A=2< \sup A =3 < \sup S = 4$.

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To do this formally just use the definitions.

Given: $S$ is bounded above and below. As $\mathbb R$ has the least upper and lower bound property. $\inf S$ and $\sup S$ exist.

Claim: $A$ is bounded above and below.

Pf: $S$ is bounded below so there is a $k$ so that $k \le s$ for every $s\in S$. For any $a\in A$, $a \in S$ because that's what a subset means. So $k\le a$ becasue $k$ is less or equal to every element of $S$. So $A$ is bounded below.

So $A$ is bounded below. THe same argument holds to show $A$ is bounded above. And by l.u.b. property $\inf A$ and $\sup A$ exist.

Claim: $\inf S \le \inf A$.

I'll leave the proof to you but it is a similar subset argument.

$\inf S$ is a lower bound of $S$. Show that means $\inf S$ is a lower bound of $A$. For any $k; k > \inf A$ then $k$ can not be a lower bound of $A$ so show it can't be a lower bound of $S$ either. What happens if $\inf S > \inf A$?

Claim

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