Bounded subsets (supremum and infimum)

Question

The problem is as follows:

Let $S$ and $T$ be nonempty, bounded subsets of the real numbers. Prove that if $T$ contains $S$, then $\inf T\leq\inf S\leq \sup S\leq\sup T.$

My problem is that I don’t understand why the parent set has the greatest lower bound and the least upper bound. Help. Please and thank you.

Answer 1

The first thing you need to know is that since the sets are nonempty and bounded, the Completeness of the Reals says that they have suprema and infima.

The main insight you need to have is that if $S\subseteq T$, then every upper (resp. lower) bound of $T$ is also an upper (resp. lower) bound of $S$. Then explain how this gives you the inequality involving the sups and infs.

Answer 2

We know that $S \subseteq T$, let's see what happens if $\inf S < \inf T$. If this were true, then $\inf T - \inf S > 0$, so there exists an $s \in S$ such that $\inf S \leq s < \inf T$ (to see this: define $\varepsilon = \inf T - \inf S > 0$ which allows us to find some $s \in S$ such that $\inf S \leq s < \inf S + \varepsilon$ and finally conclude $\inf S \leq s < \inf S + \inf T - \inf S = \inf T$).

As we know that $\inf S \leq s < \inf T$, we can state that $s$ is an strict lower bound for $T$, i.e., for all $t \in T$, $s < t$, which is a contradiction since $S \subseteq T$ and thus $s \in T$ ($\implies s<s$). We conclude that $\inf S \geq \inf T$.

By definition we know $\inf S \leq \sup S$ and using the above argument for the supremum we finally have $\inf T \leq \inf S \leq \sup S \leq \sup T$.

I hope it helps :)

Answer 3

"My problem is that I don’t understand why the parent set has the greatest lower bound and the least upper bound."

Well, both sets are bounded.

Both sets are bounded so upper and lower bounds exist for both sets.

And as they are both subsets of the Real Numbers (and neither is empty) it is the nature (actually the definition almost) of real numbers that all bounded non-empty sets do have least upper bounds and greatest lower bounds, so both the sets have a $\sup$ and an $\inf$.

The main thing to realize is that if $S\subset T$ and $u$ is greater than all elements of $T$ then $u$ is greater than all elements of $S$ because all elements of $S$ are elements of $T$.

From there if you understand the definitions, everything falls into place.

$\sup S$ must exist because $S$ is bounded above. $\sup T$ must exist because $T$ is bounded above. $\sup T \ge x$ for all elements of $T$ because $\sup T$ is an upper bound of $T$. So $\sup T \ge y$ for all elements of $S$ because if $y \in S$ then $y \in T$ because $S\subset T$. So $\sup T$ is an upper bound of $S$ (as well as for $T$). But $\sup S$ is the least upper bound so $\sup S \le u$ for all possible upper bounds of $S$. And that include $\sup T$ which is also an upper bound of $S$. So $\sup S \le \sup T$.

The same arguments can be made about $S$ and $T$ being bounded below so $\inf S$ and $\inf T$ exist and $\int T$ being a lower bound of $T$ means it is at least as small as any element of $T$ and that includes the elements of $S$ so $\int T$ is a lower bound of $S$ and as $\inf S$ it the greatest lower bound of $S$ than $\inf T \le \inf S$. So $\inf T \le \inf S$.

Just have to show $\inf S \le \sup S$.

which should be obvious. Every lower bound of $S$ is at least as small as any element of $S$, and every upper bound is at least as large as that element of $S$. so every lower bound of $S$ is smaller or equal to every upper bound of $S$. So $\inf S \le \sup S$.

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Alternatively:

$S$ is bounded above. $S$ is not empty. And $S\subset \mathbb R$. So because the reals have the least upper bound property $\sup S$ exists.

But the same reasoning $T$ is bounded above and non-empty and a subset of the realls so $\sup T$ exists.

Now either $\sup S\le \sup T$ or $\sup S > \sup T$.

Suppose $\sup s > \sup T$. Now the definition of $\sup S$ says that if anything is less than $\sup S$ then it can not be an upper bound of $S$. So if $\sup T < \sup S$ then $\sup T$ is not an upper bound of $S$. That means there is an $s \in S$ so that $\sup T < s$.

But $\sup T \ge t$ for all $t \in T$ so if $s > \sup T$ then $s \not \in T$.

And if $s \in S$ but $s \not \in T$ then $S\not \subset T$.

But we were told $S\subset T$ so we have reached a contradiction. $\sup S > \sup T$ is impossible so $\sup S \le \sup T$.

.....

$S$ is non-empty and bounded below so $\inf S$ exist. $S$ is not empty so there exists a $k\in S$. By the definition of $\sup S$ and $\inf S$ we know $\inf S \le s$ and $s \le \sup S$. And above we proved $\sup S \le \sup T$.

so $\inf S \le s \le \sup S \le \sup T$.

....

Now we repeat the argument two sections above to prove $\inf T$ exists and is $\le \inf T$.

So $\inf T\le \inf S \le s \le \sup S\le \sup T$.