Difference between revisions of "Talk:Lebesgue number lemma"
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| + | ==Questionable proof== | ||
No, I do not think you can just choose the least out of a finite number of deltas. | No, I do not think you can just choose the least out of a finite number of deltas. | ||
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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. [[User:Boris|Boris]] ([[User talk:Boris|talk]]) 17:39, 10 May 2016 (UTC) | 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. [[User:Boris|Boris]] ([[User talk:Boris|talk]]) 17:39, 10 May 2016 (UTC) | ||
| + | : Nice to hear from you again. If you take ANY point {{M|x\in X}} there exists {{M|B_{\frac{1}{2}\epsilon_{x_i} }(x_i)\ni x}}. | ||
| + | :* This means {{M|d(x,x_i)<\frac{1}{2}\epsilon_{x_i} }} | ||
| + | : If {{M|1=\delta:=\text{min}\left(\{\frac{1}{2}\epsilon_{x_i}\ \vert\ x_i\in\text{ that finite open cover}\}\right) }} then: | ||
| + | :* {{M|\forall i\in\{1,\ldots,n\}[\delta\le\frac{1}{2}\epsilon_{x_i}]}} | ||
| + | : Let {{M|A}} be a set of diameter {{M|\delta}} | ||
| + | :* Let {{M|\alpha\in A}} be arbitrary. | ||
| + | :** There exists an {{M|x_i}} such that: {{M|\alpha\in B_{\frac{1}{2}\epsilon_{x_i} }(x_i)}} | ||
| + | :*** Thus {{M|d(\alpha,x_i)<\frac{1}{2}\epsilon_{x_i} }} | ||
| + | :** By the definition of diameter of {{M|A}} we have: | ||
| + | :*** {{M|\forall\alpha,\beta\in A[d(\alpha,\beta)<\delta]}} | ||
| + | :** Let {{M|\beta\in A}} be arbitrary | ||
| + | :*** {{M|d(\beta,x_i)\le d(\beta,\alpha)+d(\alpha,x_i)<\delta+\frac{1}{2}\epsilon_{x_i} }} | ||
| + | :*** As {{M|\delta\le\frac{1}{2}\epsilon_{x_i} }} for all {{M|i}} we see: | ||
| + | :**** {{M|1=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: | ||
| + | :***** {{M|1=d(\beta,x_i)<\epsilon_{x_i} }} | ||
| + | :**** This is the very definition of {{M|\beta\in B_{\epsilon_{x_i} }(x_i)\subseteq U\in\mathcal{U} }} | ||
| + | :*** We have shown: {{M|\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. [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 21:16, 11 May 2016 (UTC) | ||
| + | |||
| + | ::Oops! | ||
| + | ::Indeed, you are right. | ||
| + | ::[[User:Boris|Boris]] ([[User talk: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. [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 08:38, 29 May 2016 (UTC) | ||
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 [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]
- [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:
- We have shown: [ilmath]\beta\in B_{\epsilon_{x_i} }(x_i)\subseteq U\in\mathcal{U} [/ilmath] - this completes the proof.
- There exists an [ilmath]x_i[/ilmath] such that: [ilmath]\alpha\in B_{\frac{1}{2}\epsilon_{x_i} }(x_i)[/ilmath]
- Let [ilmath]\alpha\in A[/ilmath] be arbitrary.
- I don't see a problem with this. Alec (talk) 21:16, 11 May 2016 (UTC)
