Uniform continuity proof on bounded sets

[lionel_hutz]

Homework Statement

Prove that if f is uniformly continuous on a bounded set S, then f is a bounded function on S.

Homework Equations

Uniform continuity: For all e>0, there exist d>0 s.t for all x,y in S |x-y| implies |f(x)-f(y)|

The Attempt at a Solution

Every time my book has covered a similar topic, it uses subsequences, which I'm a bit uncomfortable with. Is this following proof valid?

Let f be uniformly continuous on a bounded set, S (1), Then:

For all e>0, there exist d>0 s.t for all x,y in S |x-y| implies |f(x)-f(y)| (2)

so (3) |f(x)|<|f(y)| + e, for all x in S

Therefore, it's bounded on S

 

[jbunniii]

Homework Statement

Prove that if f is uniformly continuous on a bounded set S, then f is a bounded function on S.

Homework Equations

Uniform continuity: For all e>0, there exist d>0 s.t for all x,y in S |x-y| implies |f(x)-f(y)|

The Attempt at a Solution

Every time my book has covered a similar topic, it uses subsequences, which I'm a bit uncomfortable with. Is this following proof valid?

Let f be uniformly continuous on a bounded set, S (1), Then:

For all e>0, there exist d>0 s.t for all x,y in S |x-y| implies |f(x)-f(y)| (2)

so (3) |f(x)|<|f(y)| + e, for all x in S

Therefore, it's bounded on S

How did you get (3)? You don't know that $|x-y| < d$ for all $x,y \in S$.

This same question was just asked yesterday. Perhaps the hints given in that thread will help:

https://www.physicsforums.com/showthread.php?t=678514

 

[LCKurtz]

Uniform continuity: For all e>0, there exist d>0 s.t for all x,y in S |x-y| implies |f(x)-f(y)|

I thought that was just a typo until..

The Attempt at a Solution

Let f be uniformly continuous on a bounded set, S (1), Then:

For all e>0, there exist d>0 s.t for all x,y in S |x-y| implies |f(x)-f(y)| (2)
S

Until I saw you "use" it. It isn't even a sentence. I think you need to work on understanding the definition of uniform continuity.