Prove f is bounded on A using uniform continuity

[kingstrick]

Homework Statement

Prove that if F is uniformaly continuous on a bounded subset of ℝ, then F is bounded on A.

Homework Equations

The Attempt at a Solution

F is uniformaly continuous on a bounded subset on A in ℝ. Therefore each ε>0, there exists δ(ε)>0 st. if x, u is in A where |x-u|<δ(ε) then |f(x)-f(u)|<ε. Since f is uniformaly continuous on A, then it is continuous at every point of A. Since A is a bounded subset, there exists m>0 st m > |x| for all x in A. Lastly, assume f is unbounded on A. Since f is unbounded on A, if given k >0 there exists a sequence (Xk) in A where |f(Xk)|>k for all k in ℝ. Since A is bdd, seq (Xk) is bdd. Since F is continuous at every point and A is bdd. then there exist a subequence of (Xk) denoted by (Xkn) that converges to x. Then it follows that F(Xkn) converges to F(x). Therefore F(Xkn) must be bounded. This is a contradiction a this would imply that there exists k₀ where |F(Xkn)|≤ k₀ while |F(Xkn)| should be ≥ k for all k in ℝ.

**My proofs are always long winded like this**
If at all possible, could somebody first, help me if i solved this problem wrong, and if i am right, show me a more concise way to write this problem up.

 

[jgens]

F is uniformaly continuous on a bounded subset on A in ℝ. Therefore each ε>0, there exists δ(ε)>0 st. if x, u is in A where |x-u|<δ(ε) then |f(x)-f(u)|<ε. Since f is uniformaly continuous on A, then it is continuous at every point of A. Since A is a bounded subset, there exists m>0 st m > |x| for all x in A. Lastly, assume f is unbounded on A. Since f is unbounded on A, if given k >0 there exists a sequence (Xk) in A where |f(Xk)|>k for all k in ℝ. Since A is bdd, seq (Xk) is bdd.

Everything through this point is correct.

Since F is continuous at every point and A is bdd. then there exist a subequence of (Xk) denoted by (Xkn) that converges to x.

Here is where you are going to run into difficulties. The sequence $\{x_k\}_{k \in \mathbb{N}}$ converges in $\mathbb{R}$ but that does not mean it converges in $A$. To make this proof work, you would need $A$ to be a closed (and bounded) subset of $\mathbb{R}$.

Then it follows that F(Xkn) converges to F(x). Therefore F(Xkn) must be bounded. This is a contradiction a this would imply that there exists k₀ where |F(Xkn)|≤ k₀ while |F(Xkn)| should be ≥ k for all k in ℝ.

This part of the argument is contingent upon the fact that $\{x_k\}_{k \in \mathbb{N}}$ converges in $A$. Since $A$ is not closed, this does not necessarily follow.

**My proofs are always long winded like this**
If at all possible, could somebody first, help me if i solved this problem wrong, and if i am right, show me a more concise way to write this problem up.

I cannot think of any way to salvage your proof off the top of my head, but I can suggest how I would prove the result. Choose $0 < \delta$ such that $|f(x_1)-f(x_2)| < 1$ whenever $x_1,x_2 \in A$ satisfy $|x_1-x_2| < \delta$. Since $A$ is bounded, you can cover $A$ by a finite collection of sets with diameter less than $\delta$. Can you show how this forces boundedness?

 

[kingstrick]

Is this better?

F is uniformaly continuous on a bounded subset on A in ℝ. Therefore each ε>0, there exists δ(ε)>0 st. if x, u is in A where |x-u|<δ(ε) then |f(x)-f(u)|<ε. Since f is uniformaly continuous on A, then it is continuous at every point of A. Since A is a bounded subset, there exists m>0 st m > |x| for all x in A. Lastly, assume f is unbounded on A. Therefore for all k >0, there exists an x such that |f(x)| ≥k. Therefore for each n in N there is an Xn in A such that |f(Xn)|≥n. The (Xn)'s ae a sequence in the bounded set A. Therefore by the Bolzano-Weirstraus (Xn) has a convergent subsequence (Xnr). So (Xnr) is Cauchy. Thus F is uniforaml continuous at (f(Xnr)). However since |f(Xnr)|≥|f(Xn)|≥n implies (f(Xnr)) is divergent, we have a contradiction.

So F is bounded on A.

 

[jgens]

Is this better?

Unfortunately not. It has the same problem as the first argument.

Therefore by the Bolzano-Weirstraus (Xn) has a convergent subsequence (Xnr). So (Xnr) is Cauchy. Thus F is uniforaml continuous at (f(Xnr)).

This does not show uniform continuity. Based on your reasoning here, it would probably be a good idea for you to review the definition of uniform continuity.

I will suggest an alternative method of proof that is quite simple: Take $0 < \delta$ such that $|f(x_1)-f(x_2)| < 1$ whenever $x_1,x_2 \in A$ satisfy $|x_1-x_2| < \delta$. Now find a finite cover of $A$ by sets of the form $(x-2^{-1}\delta,x+2^{-1}\delta) \cap A$ (here is where the boundedness of $A$ comes into play). Now use these sets to show that $f$ is necessarily bounded on $A$.

Edit: I thought of a fairly simple way to make your argument by contradiction work. If you really want to go with that method of proof, then I can walk you through that. The idea utilized in the direct proof comes up quite a bit in analysis and properties involving finite covers come up a lot in topology, so I think understanding the direct proof is helpful. If you can't tell, I tend to have a preference for direct proofs rather than proofs by contradiction.

 

[kingstrick]

Unfortunately not. It has the same problem as the first argument.
This does not show uniform continuity. Based on your reasoning here, it would probably be a good idea for you to review the definition of uniform continuity.

I will suggest an alternative method of proof that is quite simple: Take $0 < \delta$ such that $|f(x_1)-f(x_2)| < 1$ whenever $x_1,x_2 \in A$ satisfy $|x_1-x_2| < \delta$. Now find a finite cover of $A$ by sets of the form $(x-2^{-1}\delta,x+2^{-1}\delta) \cap A$ (here is where the boundedness of $A$ comes into play). Now use these sets to show that $f$ is necessarily bounded on $A$.

Edit: I thought of a fairly simple way to make your argument by contradiction work. If you really want to go with that method of proof, then I can walk you through that. The idea utilized in the direct proof comes up quite a bit in analysis and properties involving finite covers come up a lot in topology, so I think understanding the direct proof is helpful. If you can't tell, I tend to have a preference for direct proofs rather than proofs by contradiction.

I think the proof by contradiction will be more helpful as I don't understand where the -2^-1 came frm in your neighborhoods. Please help.

 

[jgens]

I think the proof by contradiction will be more helpful as I don't understand where the -2^-1 came frm in your neighborhoods. Please help.

Suppose that $f$ is uniformly continuous on $A$ but unbounded. For each $k \in \mathbb{N}$ choose $x_k \in A$ such that $k \leq f(x_k)$. The sequence $\{x_k\}_{k \in \mathbb{N}}$ is bounded and therefore has some Cauchy subsequence $\{x_{k_n}\}_{n \in \mathbb{N}}$. Choose $0 < \delta$ such that $|f(x_1)-f(x_2)| < 1$ whenever $x_1,x_2 \in A$ satisfy $|x_1-x_2| < \delta$. Choose $N \in \mathbb{N}$ so large that $N \leq n,m$ implies $|x_{k_n}-x_{k_m}| < \delta$. Can you derive a contradiction from here?

After you get the proof using contradiction, it would really be a good idea to do the direct proof as well.

 

[kingstrick]

Is this better?

F is uniformaly continuous on a bounded subset on A in ℝ. Therefore each ε>0, there exists δ(ε)>0 st. if x, u is in A where |x-u|<δ(ε) then |f(x)-f(u)|<ε. Since f is uniformaly continuous on A, then it is continuous at every point of A. Since A is a bounded subset, there exists m>0 st m > |x| for all x in A. Lastly, assume f is unbounded on A. Therefore for all k >0, there exists an x such that |f(x)| ≥k. Therefore for each n in N there is an Xn in A such that |f(Xn)|≥n. The (Xn)'s ae a sequence in the bounded set A. Therefore by the Bolzano-Weirstraus (Xn) has a convergent subsequence (Xnr). So (Xnr) is Cauchy. Thus F is uniforaml continuous at (f(Xnr)). However since |f(Xnr)|≥|f(Xn)|≥n implies (f(Xnr)) is divergent, we have a contradiction.

So F is bounded on A.

Suppose that $f$ is uniformly continuous on $A$ but unbounded. For each $k \in \mathbb{N}$ choose $x_k \in A$ such that $k \leq f(x_k)$. The sequence $\{x_k\}_{k \in \mathbb{N}}$ is bounded and therefore has some Cauchy subsequence $\{x_{k_n}\}_{n \in \mathbb{N}}$. Choose $0 < \delta$ such that $|f(x_1)-f(x_2)| < 1$ whenever $x_1,x_2 \in A$ satisfy $|x_1-x_2| < \delta$. Choose $N \in \mathbb{N}$ so large that $N \leq n,m$ implies $|x_{k_n}-x_{k_m}| < \delta$. Can you derive a contradiction from here?

After you get the proof using contradiction, it would really be a good idea to do the direct proof as well.

Since we both derived a subsequence that is cauchy, why doesn't my solution about the cauchy sequences being divergent work as an acceptable contradiction?

 

[jgens]

Since we both derived a subsequence that is cauchy, why doesn't my solution about the cauchy sequences being divergent work as an acceptable contradiction?

Because you are missing all of the work that shows that this violates uniform continuity. Your level of proof writing is not at the point where you can ask the reader to fill in the details; for one, I am not convinced you actually know why the Cauchy criterion gives you a contradiction. You need to explicitly write out why you get a contradiction.

To my fault, I should have thought about how to use Cauchy sequences to derive a contradiction sooner and pointed you down that road (since you seem to be more comfortable with that).

 

[kingstrick]

thank you for your help. I ran out of time.