Prove T is bounded

**wopashui** · March 6th 2010, 11:00 AM

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

Show that T is bounded

**Plato** · March 6th 2010, 12:12 PM

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

$T \subseteq S \Rightarrow \quad x \in T \to x \in S$

**wopashui** · March 6th 2010, 01:16 PM

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?

**Plato** · March 6th 2010, 01:47 PM

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.

**HallsofIvy** · March 6th 2010, 02:04 PM

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.

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