Here is a proof that I am having trouble getting started with. I'd appreciate any help.
Let be a subset of . Let \rightarrow \mathbb{R}" alt="f\rightarrow \mathbb{R}" /> be uniformly continuous. Let be a limit point of . Suppose . Prove there is a continuous function \bigcup \{x_0\}\rightarrow \mathbb{R}" alt="g\bigcup \{x_0\}\rightarrow \mathbb{R}" />, such that for all .
I'll be happy to attempt the proof. If anyone could maybe just tell me exactly what direction I should be looking to go I would would appreciate it.
To be honest we haven't talked about Cauchy sequences so I'm not sure how to connect all these.
However, I will do my best with what I have gathered form Wikipedia...
Since is a limit point in then for there must be a sequence of points that are contained in that converge to .
Therefore, we have a sequence of points that are converging to a point . Thus,(from what I gather about Cauchy Sequences), this sequence is a Cauchy sequence since the points are converging to .
Then, (I got the following from Wikipedia) M is said to be complete (or Cauchy) if every Cauchy sequence of points in M has a limit that is also in M or alternatively if every Cauchy sequence in M converges in M.
Here, our Cauchy sequences is converging to where . So, is complete?
Sorry this is might be kinda weak but I did what I could for never have learning about Cauchy Sequences or completeness....
Hmm, well is complete (any closed subspace of a complete space is complete), but the point is this.
Picture this. Since there is a sequence of points in such that . Now, these points converge to but all we really care about is that they get really close to one another as gets big (they're Cauchy). But, it's a fact that if the 's get really close to one another (Cauchy again) and is unif. cont then the values of the sequence get really close to one another (Cauchy). But, it's a fact that says that all sequences of real numbers which get really close to one another (Cauchy) converge, and so in particular converges to some . So, what if we said that . I mean, it's not obvious why this construction is independent of the choice of the sequence , but it is. From there continuity (and in fact uniform continuity, although you don't need to prove that) are clear since if is any sequence in we have that (combining continuity on with the construction of ) that for any in .