91Ó°ÊÓ

The set \([0,1)\) is neither open nor closed in \((\mathbb{R},|\cdot|)\). (This shows that the ideas of open and closed are not mutually exclusive there are some sets in each normed linear space which are neither open nor closed.)

Prove that \(\mathbb{Z}\) (the integers) is a closed subset of \(\mathbb{R}\).

Let \(A\) and \(B\) be disjoint closed sets in a normed linear space \(X\). Construct a continuous function \(f: X \rightarrow[0,1]\) such that \(f(a)=0\) for all \(a\) in \(A\), and \(f(b)=1\) for all \(b\) in \(B\).

The following constructs the Cantor set in \(\mathbb{R}\) : begin with the unit interval \([0,1] .\) The set \(F_{1}\) is obtained from this set by removing the middle third, so that $$ F_{1}=\left[0, \frac{1}{3}\right] \cup\left[\frac{2}{3}, 1\right] $$ Now \(F_{2}\) is obtained from \(F_{1}\) by removing the middle third from each of the constituent intervals of \(F_{1} .\) Hence $$ F_{2}=\left[0, \frac{1}{9}\right] \cup\left[\frac{2}{9}, \frac{1}{3}\right] \cup\left[\frac{2}{3}, \frac{7}{9}\right] \cup\left[\frac{8}{9}, 1\right] $$ In general, \(F_{n}\) is the union of \(2^{n}\) intervals, each of which has the form $$ \left[\frac{k}{3^{n}}, \frac{k+1}{3^{n}}\right] $$ where \(k\) lies between 0 and \(3^{n} .\) The Cantor set \(F\) is what remains after this process has been carried out for all \(n\) in \(\mathbb{N}\). Show that (i) \(F\) is closed in \(\mathbb{R}\) (ii) int \(F\) is empty (iii) \(F\) contains no non-empty open set. (iv) the complement of \(F\) can be expressed as a countable union of open intervals. (If you are unfamiliar with the definition of a countable set, then Appendix A gives a brief introduction to this concept.)