Dgdrxrt

I want to find a set $mathcalO$ that satisfies the following definition
:
$$exists varepsilon > 0 : forall x in mathcalO ; textit is ; (x-varepsilon,x + varepsilon ) subseteq mathcalO$$

I think the only set that would work in this case would be the set of all real numbers, but I'm not sure how to go about proving it.

The set $mathcalO$ is clearly open (it is the union of the intervals $(x-varepsilon,x+varepsilon)$ for $xin mathcalO$). On the other hand, it is also closed. To see this, let $(x_n)$ be a sequence in $mathcalO$ converging to a real number $x$. Then $|x_n-x|<varepsilon$ for sufficiently large $n$. But then $x$ belongs to the interval $(x_n-varepsilon,x_n+varepsilon)$, which is contained in $mathcalO$ by hypothesis. Thus $xin mathcalO$.

Conclusion: $mathcalO$ is both open and closed. But the only such subsets of $mathbbR$ are $varnothing$ and $mathbbR$ itself.

I apologize if topology on $mathbbR$ is out of the scope of the question.

In case you want to understand the above argument, let me provide the necessary definitions.

Here are two definitions for a subset $X$ of $mathbbR$:

• 
  $X$ open if it is a union of open intervals.


• 
  $X$ is closed if its complement in $mathbbR$ is open.

It's a simple exercise to show that $X$ is closed if and only if it is closed under taking limits: if $(x_n)$ is a sequence in $X$ converging to a point $xin mathbbR$, then $xin X$.

Finally, a set is clopen if it is both closed and open. Trivially, the sets $varnothing$ and $mathbbR$ are clopen.

The topology theorem I used above is the following.

Theorem: The only clopen subsets of $mathbbR$ are $varnothing$ and $mathbbR$.

This theorem is actually notably difficult to prove, so a lot of details are swept under the rug here.

If $x notin mathcalO$, then $(x-varepsilon, x+varepsilon)$ is disjoint from $mathcalO$ (for the given $varepsilon$ as in the condition on $mathcalO$): if not, $p in mathcalO$ existed with $|x-p| < varepsilon$, but this in turn implies that $x in (p-varepsilon,p+varepsilon) subseteq mathcalO$ which is a contradiction. Hence $mathcalO$ is closed and as $mathbbR$ is connected, $mathcalO=emptyset$ or $mathcalO=mathbbR$. Both do satisfy the condition...

If $O not = emptyset$ then

Suppose $O not= mathbbR $, then there exist $y in mathbbR$ such that $y notin O$

Let $x in O$, suppose $x>y$, then there exists $delta = sups>0:(-s + x,x+varepsilon)subset O$

Then $a = -delta + x +fracvarepsilon2 in O $

Then $(a-varepsilon,a+varepsilon) subset O$

Then $(a-varepsilon,a+varepsilon)cup(x-delta,x+varepsilon) subset O$

But $(a-varepsilon,a+varepsilon)cup(x-delta,x+varepsilon) = (a-varepsilon,x+varepsilon) = (-delta -fracvarepsilon2 + x, x + varepsilon)$

This contradicts that $delta$ was the supreme

The same if $x<y$

Then $O = mathbbR$