- #1

Hodgey8806

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**Prove that the following are equivalent: a) A is bounded, b) A is "in" a closed ball**

## Homework Statement

The full problem is:

Let M be a metric space an A[itex]\subseteq[/itex]M be any subset. Prove that the following are equivalent:

a)A is bounded.

b)A is contained in some closed ball

c)A is contained in some open ball.

I only want help going from A to B, but maybe a little guidance from B to C or A to B--and I will attempt to prove the opposite way.

## Homework Equations

Book definitions:

A is bounded if [itex]\exists[/itex]R≥0 s.t. d(x,y)≤R [itex]\forall[/itex] x,y[itex]\in[/itex]A

If a is a nonempty bounded subset of M, the diameter of A is diam(A) = sup{d(x,y):x,y[itex]\in[/itex]A}

For any x[itex]\in[/itex]M and r>0, the closed ball of radius r around x is [itex]\overline{B}[/itex]

_{r}(x)={y[itex]\in[/itex]M:d(y,x)≤R}

## The Attempt at a Solution

My first thoughts are:

(=>) A to B

Spse A is bounded.

Let R = diam(A)

[itex]\forall[/itex]x

_{1},x

_{2}[itex]\in[/itex]A, d(x

_{1},x

_{2})≤R

Thus, [itex]\forall[/itex]x[itex]\in[/itex]A, [itex]\exists[/itex]y[itex]\in[/itex]M s.t. d(y,x)≤R

Let [itex]\overline{B}[/itex]

_{r}(x)={y[itex]\in[/itex]M:d(y,x)≤R} be the arbitrary union of y's.

Thus, [itex]\forall[/itex]x[itex]\in[/itex]A, x[itex]\in[/itex][itex]\overline{B}[/itex]

_{r}(x)

Thus, A[itex]\subseteq[/itex][itex]\overline{B}[/itex]

_{r}(x)