MHB Lipschitz Continuity .... and Continuity in R^n ....

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I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

In Definition 1.3.4 D&K define continuity and then go on to define Lipschitz Continuity in Example 1.3.5 ... ... (see below for these definitions ...)I have tried to show that Lipschitz Continuity implies continuity of a mapping f ... but have not succeeded ...

Can someone please demonstrate how to rigorously prove that Lipschitz continuity implies that f is continuous ... (***edit*** ... better still would be to show that Lipschitz continuity implies that f is uniformly continuous ...)Help will be appreciated ...

Peter========================================================================================***Note***

D&K's definitions of continuity and Lipschitz continuity read as follows:https://www.physicsforums.com/attachments/7749Peter
 
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Hi, Peter.

For a given epsilon, choose $\delta=\epsilon/k$. Note that the choice of $\delta$ works for all $x$, which implies uniform continuity.
 
GJA said:
Hi, Peter.

For a given epsilon, choose $\delta=\epsilon/k$. Note that the choice of $\delta$ works for all $x$, which implies uniform continuity.

Thanks GJA ... ...

I think I can see this ... since then we have ...

$$\mid \mid x - x' \mid \mid \Longrightarrow \mid \mid f(x) - f(x') \mid \mid\le k \mid \mid x - x' \mid \mid = k \delta = \epsilon
$$Is that correct?

Peter
 
You got it!
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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