## Spot the Not

This one is a riddle of my own invention. It gives a good counter-example to something that we tend to take for granted.

The riddle requires some knowledge of Topology and Real-Analysis. For those of you lacking it, all the relevant definitions are included at the end (I recommend skimming through them before reading the riddle itself).

#### This seemingly trivial list of claims leads to a contradiction. Can you find the error?

- If \(A\) is an open subset of \(\mathbb{R}\), then \(\mathbb{R}-A\) is a closed subset.
- If \(A\) is a subset of \(\mathbb{R}\), then \(bdy(A)\) = \(bdy(\mathbb{R}-A)\).
- If \(T\) is the set of all irrational numbers in the segment \([0,1]\), then \(u(A) = 1\) (\(u\) is the Lebesgue measure).
- There exists a closed set \(B\) contained in the set \(T\) (\(T\) as in 3) such that \(u(B) > 0.9\).
- If \(A\) is an open subset of \(\mathbb{R}\), then \(A\) is a countable union of open intervals.
- If \(I\) is an open interval then \(bdy(I)\) contains at most 2 points.
- From 5 and 6 it follows that if \(A\) is an open subset of \(\mathbb{R}\), then \(bdy(A)\) is countable.
- From 1, 2 and 7 it follows that if \(A\) is a closed subset of \(\mathbb{R}\) then \(bdy(A)\) is countable.
- From 8 it follows that if \(A\) is a closed set then \(u(A) = u(A-bdy(A))\).
- From 9 it follows that \(u(B-bdy(B)) > 0.9\) (where \(B\) is as in 4).
- Let \(C=B-bdy(B)\). Then \(C\) is an open set contained in \(T\) with \(u(C) > 0.9\). This is clearly impossible since the only open set contained in \(T\) is the empty set with Lebesgue measure of 0!

*Can you “Spot the Not”?*

#### Definitions Used by the Riddle

*Open set*– a set is called open if each point of the set is an interior point of the set.*Interior point*- a point \(p\) is called an interior point of a subset \(A\) of \(\mathbb{R}\) if there exists \(e>0\) such that the interval \((c-e,c+e)\) is contained in \(A\).*Closed set*– a set is called closedif the limit of every converging sequence contained in the set is also in the set. It can be shown that if \(A\) is an open set in \(\mathbb{R}\), then \(\mathbb{R}-A\) is a closed set in \(\mathbb{R}\), and vise-versa. This is actually very easy - try to prove it!*Measure of an interval*– the measure of an interval \(I=(a,b)\) is denoted \(m(I)\) and is equal to \(b-a\).*Lebesgue outer measure*– let \(A\) be a subset of \(\mathbb{R}\). The Lebesgue outer measure of \(A\) is defined as \(\inf_{}\sum{m(P_i)}\), s.t. \(Pi\) are countably many open intervals and \(A\) is contained in union of the \(Pi\)s. Those of you unfamiliar with the definition of the Lebesgue measure, can replace the occurrences of Lebesgue measure with Lebesgue outer measure in the claims above.*Boundary*– the boundary of a subset \(A\) of \(\mathbb{R}\), denoted \(bdy(A)\) is the set \(\{ x \mid \forall \epsilon>0, \exists a_1 \in A, a_2 in \mathbb{R}-A, \{a_1, a_2\} \subset (x-\epsilon,x+\epsilon) \cap A \}\).