Prove that $\inf S \le \inf A \le \sup A \le \sup S$. 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. 
 A: 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
