Show that if $A\subseteq B$, then inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$

Is my logic correct or am I missing something?

Show that if $A\subseteq B$, then inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$

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 $A$.

Since $A \subseteq B$ the following equality can be written as inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$

Solution 1:

Let's proceed by contradiction.

1) inf B $\le$ inf A

Assume, to the contrary, that inf B > inf A. Then there is some $x$ such that inf A < $x$ < inf B, but then $x \in A $ and $x \notin B$, a contradiction since $A \subseteq B$. (Also note here: If I were being super rigorous I would need to appeal to the definition of infimum to make sure that $x$ is indeed in $A$. Notice the strict inequality.) Thus, inf B $\le$ inf A

2) sup A $\le$ sup B Assume, to the contrary that sup A > sup B, then there exists some $x$ such that sup B < $x$ < sup A. However, then $x \in A$ and $x \notin B$. Again, a contradiction for the same reason as before.

3) inf A $\le$ sup A. By definition of sup and inf.

All together inf B $\le$ inf A $\le$ sup A $\le$ sup B.

Maybe someone can post a comment to help a little bit, what property of the real numbers am I using when I assume existence of $x$ like I do above? Certainly it must exist since there are no gaps in the real numbers. That way I can say that given two real numbers $a,b$ with $a <b$ then there exists $x$ such that $a <x <b$. Is this the Archimedian property? something topological?