$S\subseteq T$ implies $\inf T\leq\inf S\leq\sup S\leq \sup T$ Let $S$ and $T$ be nonempty bounded subsets of $\mathbb{R}$ with $S\subseteq T$. How do I prove that $\inf T\leq\inf S\leq\sup S\leq \sup T$?
 A: $S\subseteq T$ means every member of $S$ is a member of $T$.
$a = \inf S$ means $a$ is the largest lower bound of $S$, so it's $\le$ every member of $S$ and $\ge$ all other lower bounds of $S$.
If a number $b$ is $\le$ every member of $T$, and every member of $S$ is a member of $T$, then $b$ is $\le$ every member of $S$.  Therefore $\inf T$ is $\le$ every member of $S$.  Therefore $\inf T$ is a lower bound of $S$.  Therefore $\inf T$ is $\le$ the largest lower bound of $S$.  In other words $\inf T\le \inf S$.
A similar argument shows $\sup T\ge\sup S$.
The statement that $\inf S\le \sup S$ is true only if $S\ne\varnothing$.  If $S\ne\varnothing$, then there exists $s\in S$.  And we must then have $\inf S\le s\le\sup S$.
The fact that if $S\subseteq T$ then $\inf S\ge\inf T$ shows that $\inf\varnothing\ge$ all other "inf"s, so $\inf\varnothing=\infty$.  Similarly $\sup\varnothing=-\infty$.
A: Hint:
Direct proof:


*

*Let $(x_n)_{n \in \mathbb{N}} \subseteq S$ be a sequence such that $\lim_{n \to \infty} x_n = \inf S$.

*Since $S \subseteq T$ we have $(x_n)_{n \in \mathbb{N}} \subseteq T$ and so $\lim_{n \to \infty} x_n \geq \inf T$.


By contradiction:


*

*Assume that $\inf S < \inf T$. 

*By definition of infimum, there exists $x \in S$ such that $x < \inf T$, but $x \in T$, contradiction with the definition of $\inf T$.


I hope this helps $\ddot\smile$
A: I went about the proof this way. Please tell me if I am missing something.
Case 1. If $A \subset B$ then there exists $b_1,b_2\in B$ such that $b_1,b_2 \notin  A$. Let $a_1,a_2 \in A$ such that $a_1=$ inf A and $a_2$ = sup A. Suppose $b_1$=inf B and $b_2$=sup B than $b_1<a_1<a_2<b_2$ which implies inf $B<$ inf $A < $ sup $A<$ sup $B$
Case 2. If $A = B$ than every element in $A$ is in $B$. This implies that if $A$ is bounded above or below so is $B$ and vice versa. If the sup $B$ is defined to be the least upper bound and the inf $B$ is defined to be the greatest lowest bound. Than sup $B$ = sup $A$ and inf $B$ = inf $A$. \
1
Since $A \subseteq B$ the following equality can be written as inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$
