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Hello I have to proof the following Theorem:

Let $X$ be a compact metric space and $(U_{\alpha})$ an open cover. Then there is $\varepsilon >0$ s.t for all $A\subset X$ with $diam(X)<\varepsilon$ $A$ is contained in one element of the cover.

My idea was to use the Lebesgue Theorem which I know in the following form (Wikipedia says that this is actually the lebesgue number Theorem):

Let $X$ be a compact metric space, $Y$ a topological space covered by $(U_{\alpha})$. Let $f:X\to Y$ continuos. For all $x\in X$ there is $\varepsilon >0$ s.t $f(B_{\varepsilon}(x))$ is contained in one cover element.

Now my idea would be to chose $f=id$ but I come out always to the problem that in the second statement I know that for all $x$ there is ein epsilon s.t stomething holds for balls but in the statemend I have to proof I have to find and epsilon which works for alle sets. How can I work this out?

Paul Frost
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user1072285
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1 Answers1

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premise: I would not use the Lebesgue number Theorem in the case $f=Id$ because became a trivial statement: if $x\in X=\bigcup U_\alpha \implies x\in U_{\bar \alpha}\stackrel{U_{\bar \alpha }\text{open}}{\implies} \exists \varepsilon$ s.t. $B_\varepsilon(x)\subset U_{\bar\alpha}$.

proof: I assume by contradiction that such $\varepsilon$ doesn't exists, so for every $\varepsilon_n=1/n$ exists $A_n\subset X$ s.t. $diam(A_n)<\varepsilon_n$ and it isn't contained in any $U_\alpha$. Lets pick an element $x_n\in A_n$, now by compactness (compact$\implies $ sequentially compact) we can extract a sub-sequence $y_k=x_{n_k}$ that converges to $\bar y\in X$.

Now by the premise's (trivial) argument exists $\varepsilon >0$ s.t. $B_\varepsilon(\bar y)\subset U_{\bar\alpha}$, but now we get a contradiction because we can find such an $n$ (by the triangolare inequality) s.t.: $$A_{n_k}\subset B_{\varepsilon_k}(y_k)\subset B_\varepsilon(\bar y)\subset U_{\bar\alpha}, $$ That contradict the fact that $A_n$ isn't contained in any $U_\alpha$.

Bongo
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  • thank you, but so you never used the theorem I statet first. My Professor said that the two statements are trivially equivalent so I was wondering if there is a quick way using one another. – user1072285 Aug 10 '23 at 19:58
  • Now i don't see that equivalence, but in the premise part I said that is useless the Lebesgue number Theorem in the case $f=Id$ (because the thesis is trivially true, even if $X$ isn't metric and compact) however here i think $f=Id$ its the only possible choice to apply that theorem. – Bongo Aug 10 '23 at 22:29
  • Do you think it is possible to proof the version of this theorem with the map $f$ with the theorem you proofed ? – user1072285 Aug 11 '23 at 15:29
  • yes! if $(U_\alpha)$ is an open cover, then (by $f$'s continuity) $(V_\alpha)=(f^{-1}(U_\alpha))$ is an open cover of $X$ that is a metric and compact space. Hence, by the previous proof, there exists an $\varepsilon>0$ as in the first statement. Now for all $x\in X$, $B_{\varepsilon/2}(x)\subset V_\alpha$ (for some $\alpha$), namely $f(B_{\varepsilon/2}(x))\subset U_\alpha$. – Bongo Aug 11 '23 at 15:55
  • Ah yeah you are right thank you! – user1072285 Aug 11 '23 at 16:50