# Exercise 7.59#

For `1.`

, you can see \(U = [-∞,0) ∪ (0,∞]\). Then the complement \(U^C = [∞,0] ∪ [0,∞]\) is clearly \([0]\) or \(\{0\}\) (a point). As expected, the complement of an open set is a closed set.

The Interior (topology) of [0] is not “(0)” because an open set (an “open line”) on the real line is defined to include all points within ε of 0, where ε is non-zero (see also Ball (mathematics)). It may be best not to even think of “(0)” or “(0,0)” as an open set; these use interval notation but without two or without two different numbers to define the interval. This contrasts with the acceptable notation [0] or [0,0], but so be it. So the largest open set contained in [0] is ∅.

If we were considering the discrete topology on ℝ (rather than the standard topology), then we would consider the point {0} both open and closed.

For `2.`

the complement of ∅ is the full set. The interior of the full set is the full set.

The primary lesson here is that the pseudocomplement of the pseudocomplement of \(U\) is not always equal to \(U\).

We’ve also selected a \(U\) that is not a Regular open set. That is, it is not equal to the interior of its closure.

For `3.`

, yes. For `4.`

, no. From Heyting algebra:

Although the negation operation ¬

ais not part of the definition, it is definable asa→ 0. The intuitive content of ¬ais the proposition that to assumeawould lead to a contradiction. The definition implies thata∧ ¬a= 0. It can further be shown thata≤ ¬¬a, although the converse, ¬¬a≤a, is not true in general, that is, double negation elimination does not hold in general in a Heyting algebra.