is that function not uniformly continuous

[kavlas]

Given:

1) E is a subset of real Nos not closed

 

2) a ,is an accumulation point of E not belonging to E

 

3) f is a function of E to R (=real Nos) ,where f(x)= 1/(x-a) and ,x ,belongs to E

 

Then prove that f is not uniformly continuous in E

 

The following proof was suggested ,but i am not quite sure about it

 

Proof:

 

Let be ε>0 given. Suppose the function IS uniformly continuous (seeking a contradiction). Then there exists δ>0 such that for all x,y belonging to the real Nos with |x-y|<δ , we get |f(x)-f(y)|<ε . (The last bit being less than ,ε, is what we are going to contradict.)

 

In particular, we can pick ε = 1/2 and then there exists a δ>0 such that the above holds. WLOG, let a=0 . Then take [math] x_{n}=\frac{1}{n}[/math] and [math]y_{n}=\frac{2}{n}[/math] . With this, there certainly exists a natural No k such that for [math]n\geq k[/math], we get [math]|x_{n}-y_{n}|<\delta[/math] . But then notice the following (we are assuming a=0 , so f(x)=1/x ).

[math]|f(x_{n})-f(y_{n})| = |\frac{1}{x_{n}}-\frac{1}{y_{n}}|[/math][math] = |n-\frac{n}{2}|=\frac{n}{2}\geq\frac{1}{2}=\epsilon[/math]

 

Contradicting the uniform continuity assumption

,

[mathematic]

Looks fine to me.

[kavlas]

Looks fine to me.

 

I don't know ,man, i tried to find any theorem connecting sequences with uniform continuity and i could not.The only theorem i found was the well known one connecting simple continuity with sequences.

 

And that contradiction part it is not so clear to me.

 

What do you say

[DrRocket]

I don't know ,man, i tried to find any theorem connecting sequences with uniform continuity and i could not.The only theorem i found was the well known one connecting simple continuity with sequences.

 

And that contradiction part it is not so clear to me.

 

What do you say

 

Your proof almost works.

 

Uniform continuity doesn't have much to do with sequences. What you did was not really to use sequences but rather found a way to find two points arbitrarily close to 0 such that the value of your function at those points was arbitrarily far apart -- and that would show that the function is not uniformly continuous on an interval near 0 -- except for the problem noted below..

 

What causes the proof to not be completely valid is that the set E is somewhat arbitary and need not be an open interval with 0 as a boundary point. All that you are given is that a (which you can assume to be 0 with very little work) is an accumulation point but not an element in E. Unfortunately that means that your points x_n and y_n may not lie in E and hence may not lie in the domain of f.

 

You have the right idea, but you need a bit more finesse.

[kavlas]

 

 

What causes the proof to not be completely valid is that the set E is somewhat arbitary and need not be an open interval with 0 as a boundary point. All that you are given is that a (which you can assume to be 0 with very little work) is an accumulation point but not an element in E. Unfortunately that means that your points x_n and y_n may not lie in E and hence may not lie in the domain of f.

 

You have the right idea, but you need a bit more finesse.

 

You mean that the problem ,apart from the proof, is not correct. Because let us suppose that:

 

E ={1/n : nεN},then the 2nd sequence [math]y_{n}=\frac{2}{n}[/math] does not lie in E .

 

So unless we specify E the problem is not provable.

 

The proof is not mine ,it was suggested to me ,as i noted in the OP

[DrRocket]

You mean that the problem ,apart from the proof, is not correct. Because let us suppose that:

 

E ={1/n : nεN},then the 2nd sequence [math]y_{n}=\frac{2}{n}[/math] does not lie in E .

 

So unless we specify E the problem is not provable.

 

The proof is not mine ,it was suggested to me ,as i noted in the OP

 

No, the theorem is very provable -- using the knowledge of the properties of E given in your problem statement.

 

But you did not prove the theorem as it was stated. You proved a very special case involving some additional assumptions on E.

 

You can reduce the problem, as you did, to the case a=0 by noting that the theorem is invariant under translation. Then the critical observation is that the function f is unbounded on the intersection of E with any neighborhood of 0 and from that you can then show that f cannot be uniformly continuous since one can find two point in any such neighborhood such that the difference in the values of f on those points is arbitrarily large.

[kavlas]

No, the theorem is very provable -- using the knowledge of the properties of E given in your problem statement.

 

But you did not prove the theorem as it was stated. You proved a very special case involving some additional assumptions on E.

 

.

 

Apart from the assumption that a=0,i am sorry ,but i cannot see any other assumptions for E.

 

Please ,explain

[DrRocket]

Apart from the assumption that a=0,i am sorry ,but i cannot see any other assumptions for E.

 

Please ,explain

 

Read my earlier post.