# Simple question on continuity

1. Oct 27, 2007

### rumjum

1. The problem statement, all variables and given/known data
If X is bounded non empty subset in R (usual) and f:X->R is uniformly continuous
function. Prove that f is bounded.

2. Relevant equations

3. 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.

2. Oct 28, 2007

### zhentil

Hint: it's WAY easier than you're making it.

3. Oct 28, 2007

### rumjum

Well, how is this solution then.

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

Since X is bounded in R, it has a supremum and infimum.

Also, we can have a sequence {xm} ->x (limit point), such that x is the supremum.
Similarly, we have another sequence {x'n} such that it tends to another
limit point x' (infimum).

Let the sequence {xm} be such that x0<x1<...xm and d(xi,xj) < delta/m. Let
xm -> x and x0->x'.

Hence,
d(x0,xm) < m*delta/m = delta. In other words, the distance between
any two points of the set is always less than delta.

Then "f" is continuous uniformly and so,
d(f(xi), f(xj)) < e/m (some e >0). if d(xi,xj) < delta/m

Hence,d(f(xm),f(x0)) < d(f(xm),f(xm-1)) + d(f(xm-1),f(x0))
< e/m + d(f(xm-1,f(xm-2)) + d(f(xm-2,f(x0))..
< m*e/m = e.

Since for all points in X , the distance between the function is always
bounded, the function is bounded.

does that sound right?

4. Oct 29, 2007

### zhentil

Ok, think of the easier case: if X = [a,b] for some a and b, you could do it, right? You're not supposed to think of these functions as being differentiable, but it's ok to cheat for awhile in order to build intuition. What if X = [a,b] and f(x)=x? How would you prove that f is bounded in that case? The proof for the general case is not much worse than that.

Ok, so what if X is not nice like [a,b], then what? Well, you have to be more careful and consider various cases, but it's not too much worse.

Two comments about your solution, though: first, a bounded subset of R need not have a limit point in R. And second, sequences are infinite. And third, taking an arbitrary sequence and considering the value of the function at that point requires that the function be defined at that point; something that we can't assume.