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

Question

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 $B$

Answer 1

Your logic is incorrect in Case 1: You've assumed that $b_1 = \inf{B} \in B \setminus A$. Take $A = (0, 1)$ and $B = (0, 2)$ to see why this is quite problematic.

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For a hint towards the correct direction, can you show that any lower bound of $B$ is a lower bound of $A$ as well? So the greatest lower bound of $B$ is still a lower bound for $A$?

Answer 2

let's suppose A ⊆ B⊆R bounded then we can say that through the known theorem for real numbers Completeness Property of R: for every a ∈ A : infA ≤ a ≤ supA(1) likewise for every b ∈ B : infB ≤ b ≤ supB

but at the same time A is the subset of B so we can say that every element of A is also bounded by the least upper bound and the greatest lower bound of B for every a ∈ A : infB ≤ a ≤ supB

but from the definition of the supremum and infimum we know that for random upper bound x and lower bound y the supremum and infimum will always be smaller and greater likewise for every a ∈ A : supA ≤x and y≤ infA (2) in our problem we can assume the random bounds of x an y as x=supB , y=infB

Through (1) and (2) we have our solution inf B ≤ inf A ≤ sup A ≤ sup B