Prove T is bounded

• March 6th 2010, 11:00 AM
wopashui
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
• March 6th 2010, 12:12 PM
Plato
Quote:

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

$T \subseteq S \Rightarrow \quad x \in T \to x \in S$
• March 6th 2010, 01:16 PM
wopashui
Quote:

Originally Posted by Plato
$T \subseteq S \Rightarrow \quad x \in T \to x \in S$

how does this mean T is boundness, bounded above or below?
• March 6th 2010, 01:47 PM
Plato
The statement that $S$ is a bounded set of real numbers means that $\left( {\exists B > 0} \right)\left[ {\left( {\forall z \in S} \right) \Rightarrow \left| z \right| \leqslant B} \right]$.
Therefore $T \subseteq S\,\& \,x \in T \Rightarrow \quad \left| x \right| \leqslant B$.
That shows that $T$ is bounded.
• March 6th 2010, 02:04 PM
HallsofIvy
Quote:

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.