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For $a<b, a,b\in\Bbb R$

$$(a,b)=\bigcup_{0<\delta<(b-a)/2} I_{\delta} \quad I_{\delta}:=[a+\delta,b-\delta] $$

Clearly the RHS is an (uncountable) infinite sum of closed intervals. I have no idea how to show it is open at two ends.

(My hope is that if this is true, then it is trivial that:

$$f(x)\in\mathscr C^r(a,b)\Longleftrightarrow f(x)\in\mathscr C^r[a+\delta,b-\delta]\quad \forall \delta\in(0,\frac{b-a}2) $$

which will be very useful for me.

A proof that is neat (without too much set theoretical jargon) and based on the original definition will be sincerely appreciated. Thanks in advance!

EDIT. Thanks for Surb's answer. Now I understand why LHS and RHS are equal. But I'm still confused about another question:

Since they two are equal, doesn't it imply that if a statement A is true on LHS then it is also true on RHS? However, there are many counter-examples. Like uniform continuity for a function, or uniform convergence for a functional sequence on RHS doesn't imply that on LHS, but... How come?

Asaf Karagila
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Vim
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1 Answers1

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if $x\in (a,b)$ there is a $\delta>0$ such that $x\in[a+\delta,b-\delta]$ and thus $(a,b)\subset \bigcup_{\delta}I_\delta$. The other inclusion is obvious.

Surb
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  • Thanks for the neat proof! By the way, would you mind taking a look at my implication about $\mathscr C^r$, is that correct too? – Vim Jun 30 '15 at 12:52
  • yes, it's correct. – Surb Jun 30 '15 at 12:57
  • If you don't mind, I still have another question. Since they two are equal, doesn't it imply that if a statement A is true on LHS then it is also true on RHS? However, there are many counter-examples. Like uniform continuity for a function, or uniform convergence for a functional sequence on RHS doesn't imply that on LHS, but... How come? – Vim Jun 30 '15 at 13:11
  • Those aren't counterexamples. Note that you are not writing anything like "open interval = closed interval" -- the infinite union of closed intervals you're taking turns out to be open! The LHS and RHS are both a plain old open interval. – Lynn Jun 30 '15 at 13:35
  • @Mauris Thank you. I have get that straight. The problem lies in the "uniformity" which requires not only continuity/convergence but also a uniform threshold $N$. – Vim Jun 30 '15 at 15:37