Difference between revisions of "Talk:Lebesgue number lemma"

Latest revision as of 08:38, 29 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 $x\in X$ there exists $B_{\frac{1}{2}\epsilon_{x_i} }(x_i)\ni x$.

• This means $d(x,x_i)<\frac{1}{2}\epsilon_{x_i}$

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

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

Let $A$ be a set of diameter $\delta$

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

No problem! I'd rather a false-negative "accusation" (I can't think of a weaker word) than a false-positive slip through. Alec (talk) 08:38, 29 May 2016 (UTC)