Difference between revisions of "Talk:Lebesgue number lemma"

Revision as of 20:52, 27 May 2016

Questionable proof

No, I do not think you can just choose the least out of a finite number of deltas.

Even if the covering is finite, still, some more effort is needed; and compactness must be used again.

One option: assume the opposite; take an infinite sequence of points that are worse and worse (that is, their relevant neighborhoods are smaller and smaller); by compactness, there exist an accumulation point of this sequence; now find a contradiction...

Another option: the maximal radius of a "good" neighborhood is a function of a point, and this function is continuous (and moreover, Lipschitz(1), think why); by compactness, it has the least value. Boris (talk) 17:39, 10 May 2016 (UTC)

Nice to hear from you again. If you take ANY point [ilmath]x\in X[/ilmath] there exists [ilmath]B_{\frac{1}{2}\epsilon_{x_i} }(x_i)\ni x[/ilmath].

• This means [ilmath]d(x,x_i)<\frac{1}{2}\epsilon_{x_i} [/ilmath]

If [ilmath]\delta:=\text{min}\left(\{\frac{1}{2}\epsilon_{x_i}\ \vert\ x_i\in\text{ that finite open cover}\}\right)[/ilmath] then:

• [ilmath]\forall i\in\{1,\ldots,n\}[\delta\le\frac{1}{2}\epsilon_{x_i}][/ilmath]

Let [ilmath]A[/ilmath] be a set of diameter [ilmath]\delta[/ilmath]

• Let [ilmath]\alpha\in A[/ilmath] be arbitrary.
  • There exists an [ilmath]x_i[/ilmath] such that: [ilmath]\alpha\in B_{\frac{1}{2}\epsilon_{x_i} }(x_i)[/ilmath]
    • Thus [ilmath]d(\alpha,x_i)<\frac{1}{2}\epsilon_{x_i} [/ilmath]
  • By the definition of diameter of [ilmath]A[/ilmath] we have:
    • [ilmath]\forall\alpha,\beta\in A[d(\alpha,\beta)<\delta][/ilmath]
  • Let [ilmath]\beta\in A[/ilmath] be arbitrary
    • [ilmath]d(\beta,x_i)\le d(\beta,\alpha)+d(\alpha,x_i)<\delta+\frac{1}{2}\epsilon_{x_i} [/ilmath]
    • As [ilmath]\delta\le\frac{1}{2}\epsilon_{x_i} [/ilmath] for all [ilmath]i[/ilmath] we see:
      • [ilmath]d(\beta,x_i)\le d(\beta,\alpha)+d(\alpha,x_i)<\delta+\frac{1}{2}\epsilon_{x_i} \le \frac{1}{2}\epsilon_{x_i}+\frac{1}{2}\epsilon_{x_i}=\epsilon_{x_i}[/ilmath], or more simply:
        • [ilmath]d(\beta,x_i)<\epsilon_{x_i}[/ilmath]
      • This is the very definition of [ilmath]\beta\in B_{\epsilon_{x_i} }(x_i)\subseteq U\in\mathcal{U} [/ilmath]
    • We have shown: [ilmath]\beta\in B_{\epsilon_{x_i} }(x_i)\subseteq U\in\mathcal{U} [/ilmath] - this completes the proof.

I don't see a problem with this. Alec (talk) 21:16, 11 May 2016 (UTC)

Oops!

Indeed, you are right.

Boris (talk) 20:52, 27 May 2016 (UTC)