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
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Assume by contradiction that such limit doesn't exist.
-
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
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Since {$x_n$} approaches $x_0$, the sequence is Cauchy
Thus sequence {f($x_n$)} is also Cauchy -
Then since Cauchy sequences are convergent, the sequence {f($x_n$)} converges to some N in D
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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 ^_^
Best Answer
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:
Note that this limit has to be the same for every such sequence. So you still have to show uniqueness.
Alternatively you can show:
Do this by first showing that:
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.