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I am not student in Mathematics.

The following proof comes from the book: Real Analysis, by Elias and Rami (p.18):

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The folloing is my problem:

  1. It seems the yellow part implies the result (but it does not prove anything). What is the difference between the yellow part and the conclusion $d(K,F)\geq\delta>0$? (Hope to know the logic of this proof)
  2. If no the condition "$K$ is compact", just "$K$ is closed", then there may not exist $\delta_x>0$ such that $d(x,F)>3\delta_x$. Correct?
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sleeve chen
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  • I am not sure I understand your second problem. Could you clarify it? – Kolmin Sep 29 '15 at 20:51
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    That $3\delta_x$ exists is due to the fact that $x\in K$ implies $x\notin F$ by the assumption that the sets are disjoint, and since $F$ is closed its complement is open, so we may find an open ball around $x$ disjoint from $F$. The condition that $K$ is compact is only used to reduce the cover to a finite subcover. – String Sep 29 '15 at 20:54
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    About the first problem, it is a common procedure in these proofs: the point is that here you can take whatever $\delta_x$, so in particular you pick one (in this case $3\delta_x$) that becomes handy to establish the conclusion. – Kolmin Sep 29 '15 at 20:55
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    BTW, the finiteness of the subcover is relevant in order to be able to ensure that $\min(\delta_1,...,\delta_ N)$ is greater than zero. For infinitely many $\delta_x>0$ we would have $\inf({\delta_x})$ which could possibly be zero. – String Sep 29 '15 at 21:00
  • @Kolmin , clearly speaking, can I use the yellow part to prove $d(F,K)>0$? – sleeve chen Sep 29 '15 at 23:46
  • What I will say so is because the yellow statement says for each $x\in K$, $d(x,F)>3\delta_x$. So $d(F,K)>0$? Maybe I am not quite familiar with the way of proof in real analysis. – sleeve chen Sep 29 '15 at 23:47
  • The second problem just says if I change the condition to "$K$ is closed" (may not be compact). Then $\delta_x$ may not exist? – sleeve chen Sep 29 '15 at 23:53
  • d(x,F)>3δx>δx>0 for each x∈K, hence d(K,F)>0? Why are you not done when you show δx>0 exists via line 1? Is the proof done with just the assertion that "F is closed" and "F and K are disjoint"? – ililil Sep 29 '15 at 23:54
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    Example that the distance between 2 closed disjoint sets can be 0 is here. – ililil Sep 30 '15 at 00:22
  • @ililil so just confirm your answer. Your answer is that to the line 1(yellow part), we still have to show $\delta_x$ indeed $>0$? – sleeve chen Sep 30 '15 at 00:29
  • Similar question here with solution. – ililil Sep 30 '15 at 00:52
  • @sleevechen "the yellow statement says for each $x∈K, d(x,F)>3δ_x$ . So $d(F,K)>0d(F,K)>0$?" No, because the definition of $d(A,B)$ is $inf{d(x,y)|x∈A, y∈B}$ while $d(x,B)> \delta_x$ for each $x \in x$ only means $inf{d(x,y)|y \in B}> \delta_x$ for each $x \in A$ so $d(A,B)=inf{d(x,y)|x∈A, y∈B}=inf{d(x,B)|x∈A}$ could still be zero. – N Unnikrishnan Jan 12 '16 at 19:32

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