# Thread: Prove T is bounded

1. ## Prove T is bounded

Suppose that S is a nonempty bounded set of real numbers and T is a nonempty subset of S.

Show that T is bounded

2. Originally Posted by wopashui
Suppose that S is a nonempty bounded set of real numbers and T is a nonempty subset of S. Show that T is bounded
$\displaystyle T \subseteq S \Rightarrow \quad x \in T \to x \in S$

3. Originally Posted by Plato
$\displaystyle T \subseteq S \Rightarrow \quad x \in T \to x \in S$
how does this mean T is boundness, bounded above or below?

4. The statement that $\displaystyle S$ is a bounded set of real numbers means that $\displaystyle \left( {\exists B > 0} \right)\left[ {\left( {\forall z \in S} \right) \Rightarrow \left| z \right| \leqslant B} \right]$.
Therefore $\displaystyle T \subseteq S\,\& \,x \in T \Rightarrow \quad \left| x \right| \leqslant B$.
That shows that $\displaystyle T$ is bounded.

5. Originally Posted by wopashui
how does this mean T is boundness, bounded above or below?
Saying that a set of real numbers is "bounded" means it has both upper and lower bounds.