# An open interval as a union of closed intervals

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?

• Never mind, you changed your question ;-) – Stefan Mesken Jun 30 '15 at 12:47
• @Stefan Well.. Shoulda been \bigcup. Fixed now :) – Vim Jun 30 '15 at 12:50

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.
• 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
• @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