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The question is as follows:

Given:

(a) f is uniformly continuous on a subset D of $\mathbb R^n$

and (b) $x_0$ is a cluster point of D

Show: The limit of f(x), as x approaches $x_0$, exists in D

Here are my attempts:

Attempt 1: Direct proof

1/ By definition of uniform continuity, f is uniformly continuous on D

<==> for any $\epsilon$ > 0, there exists $\delta$ such that for any $x_n$ and $y_n$ in D,

we have d($x_n$,$y_n$) < $\delta$ ==> d(f($x_n$), f($y_n$)) < $\epsilon$

2/ By definition of a cluster point, $x_0$ is a cluster point of D ==> let N be a neighborhood about $x_0$, this neighborhood intersects D by at least one point k in D and k is different from $x_0$

Then.... I don't know how to proceed >_< I had a feeling that I have to use uniform continuity (thus continuity) to say for any x in the intersection of the neighborhood N about $x_0$ and D, f(x) should be in the neighborhood M about $f(x_0)$ But... I'm not sure.

Attempt 2: Proof by Contradiction

  1. Assume by contradiction that such limit doesn't exist.

  2. Then by negating the definition of limit, I claim there is a sequence {$x_n$} such that {$x_n$} approaches $x_0$, but {f($x_n$)} doesn't approach to any limit value L in D

  3. Since {$x_n$} approaches $x_0$, the sequence is Cauchy Thus sequence {f($x_n$)} is also Cauchy

  4. Then since Cauchy sequences are convergent, the sequence {f($x_n$)} converges to some N in D

  5. But this is not true since I claim no such limit N exists.
    Thus the original conclusion should be true.

Would someone please help me on this problem?

Thank you very much ^_^

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And where did you use uniform continuity? And what happens if $f(x_n)$ converges, but to a different limit than $f(y_n)$ for another sequence $(y_n)$ converging to $x_0$? –  k.stm May 5 '13 at 6:33
    
@K.Stm. I'm sorry but would you please clarify your idea a bit more? I added my original attempt, which I tried to use the definition of uniform continuity and cluster point, but I got stuck >_< –  Cecile May 5 '13 at 7:16
    
I just posted an answer, but I think it’s superflous, as your proof might be completely correct. I didn’t realize that if there is a sequence $(x_n)$ converging to $x_0$ such that $(f(x_n))$ doesn’t converge to $L$ while the image of another sequence $(y_n)$ converging to $x_0$ does, you can of course construct a sequence $(z_n)$ converging to $x_0$ such that $(f(z_n))$ doesn’t converge at all by mixing them. –  k.stm May 5 '13 at 7:23

1 Answer 1

up vote 1 down vote accepted

What is your codomain? Let’s say it’s $R^n$ and let $f \colon D → R^n$ be uniformly continuous.

You have to show that there is a limit $L ∈ R^n$ such that $f(x) \overset{x→x_0}\longrightarrow L$, that is:

For every sequence $(x_n)_{n ∈ ℕ}$ in $D$ converging to $x_0$ you have $f(x_n) \overset{n → ∞}\longrightarrow L$.

Note that this limit has to be the same for every such sequence. So you still have to show uniqueness.

Alternatively you can show:

(1) For every $ε > 0$ there is a $δ > 0$ such that $f(U_δ(x_0)) ⊂ U_ε(L)$.

Do this by first showing that:

(2) For every $ε > 0$ and $δ > 0$ there is a $x_{ε,δ} ∈ U_δ(x_0)$ such that $f(x_{ε,δ}) ∈ U_ε (L)$.

This you have already done. It is the choice of your $L$: The crucial point here is that, since $x_0$ is a limit point, you have a sequence in $D$ converging to $x_0$ whose image sequence still converges. You should carry out your argument that the image of that sequence is again Cauchy, using uniform continuity. Then, by completeness you get your limit $L$.

Next, show (1): Let $ε > 0$. Set $η = ε/2$ and choose $δ > 0$ such that for any $x, x' ∈ D$ one has $|f(x') - f(x)| < η$ whenever $|x - x'| < δ$. Now, for any $x ∈ U_δ(x_0)$ you have $|f(x) - L| ≤ |f(x) - f(x_{η,δ})| + |f(x_{η,δ}) - L| < η + η = ε$ and you’re done.

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