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## Homework Statement

If X is bounded non empty subset in R (usual) and f:X->R is uniformly continuous

function. Prove that f is bounded.

## Homework Equations

## The Attempt at a Solution

Since X is bounded in R, it is a subset of cell. And all cells in R

are compact.All bounded sub sets of compact sets have a limit point in R.

Hence, X has a limit point,xo in R.

But, that means a sequence in X {tn} -> x0 as n tends to infinity.

and so as t-> x0, f(t) -> q ( where q belongs to R). This implies that

we shall have some e'>0 such that dy(f(t), q) < e' , if there is some delta' >0

such that d(t,xo) < delta' . --------- (1)

Now, f is uniformly continuous, and

so we have dy(f(x), f(t)) < e/2 , for all e >0 , such that we have a delta

dx(x,t) < delta/2 (greater than zero) for all points t in X.

Now, dx(t,xo) < d(x,t) + d(x0,x)

=> dx(t,xo) < delta or some positive number M >0 --- (2)

Why? Because If x0 belongs to X then d(x,xo) < delta/2 and if xo

does not belong to X, then since X is bounded , we have some M- delta/2 such that

d(x,xo) < M -delta/2.

From (1) and (2) we have, dy(f(t),q) < e' (let's fix e'=e/2).

Is this step ok?Can we do this?

If X is open, then it is possible that the limit point xo does

not belong to X.

Now, dy(f(x),q) < dy(f(x),f(t)) + dy(f(t),q)

< e/2 + e/2 = e

But, since X is in R , it is ordered and since it is bounded, it has a

supremum. If x0 does not belong to X and is a limit point, then

xo = either inf(x) or sup(x). In any event,say if x < x0 for all x belongs to X.

Then we have |f(x)- q| to be less than some e >0. Similarly if x0< x for

all x, then |f(x)- q| to be less than some e >0. This implies that

f(x) < +sqrt((q + e)^2), where f(x) belongs to the range of f.

In other words, the range of f is always less than some positive number in R.

I am kind of lost now. Is this sufficient proof that the range is bounded.

Please help.