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Similar question: An open interval as a union of closed intervals

My approach to this question feels ham-fisted. I would appreciate,

  1. Correction if supplied proof is wrong
  2. Discussion to allieviate ham-fistedness
  3. Avoidance of the word 'clearly'

To Prove: For every $x$ in an open interval $I$, there exists a closed interval $\bar{I}_{\frac{\delta}{2}}$ strictly contained in $I$, with $x \in \bar{I}_{\frac{\delta}{2}}$.

(Here we consider $(\mathbb{R},|\cdot|)$ where the $|\cdot|$ is the Euclidean metric.)

Definition:

An interval $I\subset \mathbb{R}$ is open iff \begin{align*} \exists \delta >0, \delta \in \mathbb{R}, \text{ such that } (x-\delta,x+\delta) \subset I \quad \forall x \in I \end{align*}

Proof:

Suppose $I$ is open and $I=(a,b)$ with $a<b$, and $a,b \in \mathbb{R}$. Then pick an $x' \in I$. Since $I$ open, $\exists \delta>0$ with $(x'-\delta,x'+\delta)=I_\delta \subset I$.

But then $(x'-\frac{\delta}{2},x'+\frac{\delta}{2})=I_{\frac{\delta}{2}} \subset I_\delta \subset I$, is also an open set.

Claim: $\bar{I}_{\frac{\delta}{2}}=[x'-\frac{\delta}{2},x'+\frac{\delta}{2}]$ is a closed set and strictly contained in $I$.

We prove this by checking that $\bar{I}^c=(-\infty,x'-\frac{\delta}{2})\cup(x'+\frac{\delta}{2},+\infty)$ is an open set.

Suppose $y \in \bar{I}^c$. Then either $y \in (-\infty,x'-\frac{\delta}{2})$ or $y \in(x'+\frac{\delta}{2},+\infty) $. Let $y \in(x'+\frac{\delta}{2},+\infty) $.

Define: $\delta_1< |y-(x'+\frac{\delta}{2})|$.

Then $(x'+\frac{\delta}{2})<y<\delta_1+(x'+\frac{\delta}{2})$ and so $y \in \Big((x'+\frac{\delta}{2}),\delta_1+(x'+\frac{\delta}{2})\Big)\subset \Big((x'+\frac{\delta}{2}),+\infty \Big)$, so $\Big((x'+\frac{\delta}{2}),+\infty \Big)$ is open.

A similar argument holds for $\Big(-\infty,(x'-\frac{\delta}{2})\Big)$. Hence $\bar{I}_{\frac{\delta}{2}}\subset I$ is a closed set. Moreover it is an interval , since the closure of an open interval is again an interval.

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You’re working much too hard. Let $I=(a,b)$ be an open interval, and let $x\in I$. Let $a'=\frac12(a+x)$ and $b'=\frac12(x+b)$; then

$$a<a'<x<b'<b\;,\tag{1}$$

since

$$a=\frac{a}2+\frac{a}2<\underbrace{\frac{a}2+\frac{x}2}_{a'}<\underbrace{\frac{x}2+\frac{x}2}_x<\underbrace{\frac{x}2+\frac{b}2}_{b'}<\frac{b}2+\frac{b}2=b\;.$$

By definition $[a',b']$ is a closed interval, and it follows from $(1)$ that $x\in[a',b']\subseteq(a,b)=I$.

Some points to note:

  • It’s not necessary to choose a closed interval that has $x$ at its centre. You can do so, but it’s extra work. If you do choose to do so, one way is to let $\epsilon=\min\{x-a,b-x\}$ and let $\delta$ be any positive real number less than $\epsilon$; then $[x-\delta,x+\delta]$ is a closed interval centred at $x$ and contained in $(a,b)$, since $$a\le x-\epsilon<x-\delta<x<x+\delta<x+\epsilon\le b\;.$$

  • It’s not necessary to show that $[a',b']$ is a closed set in the topology on $\Bbb R$: you were simply asked for a closed interval, meaning a subset of $\Bbb R$ of the form $[u,v]$ for some $u,v\in\Bbb R$ with $u\le v$. Of course it happens to be true that every closed interval is a closed set in the usual topology on $\Bbb R$, but that’s a separate issue.

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  • $\begingroup$ Thanks, very nice answer. In my proof I wasn't assuming $x'$ in the center of $I$, but I wanted to put two open intervals around $x'$ so that there was 'space' to ensure a closed set 'between' them. However, as you pointed out there is clearly a more elegant approach. :) $\endgroup$ – zornslemmings Nov 4 '16 at 19:45
  • $\begingroup$ @zornslemmings: You’re welcome. (I like that screen name. :-)) $\endgroup$ – Brian M. Scott Nov 4 '16 at 19:51
  • $\begingroup$ One might also just note that $[\frac34a+\frac14b,\frac14a+\frac34b]\subset(a,b)$. +1 for a nice answer. $\endgroup$ – MPW Nov 4 '16 at 19:51

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