Chapter 3

Show that $$ \operatorname{int}(\operatorname{int} A)=\operatorname{int} A \quad \text { and } \quad \operatorname{int}(A \cap B)=\operatorname{int} A \cap \operatorname{int} B $$ Give an example to show that \(\operatorname{int}(A \cup B)=\operatorname{int} A \cup \operatorname{int} B\) is not necessarily true.

The set \(\\{x:\|x-a\| \leq r\\}\) is closed in a normed linear space \(X\).

The set \(\\{x\\}\), which consists of a single point in the normed linear space \(X\), is closed.

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 \(\mathbb{R}\).

If \(A=\\{x:\|x\|<1\\}\) and \(B=\\{x:\|x\| \leq 1\\}\) show that \(\bar{A}=B\)

Show that \(\bigcap_{n=1}^{\infty}\left(-\frac{1}{n}, 1+\frac{1}{n}\right)\) is closed in \(\mathbb{R}\).

Show that \(\bigcup_{n=2}^{\infty}\left[\frac{1}{n}, 1-\frac{1}{n}\right]\) is open in \(\mathbb{R}\).

Let \(f\) be a continuous, real-valued function on \(\mathbb{R}\). Show that the set \(\\{(a, b): b \leq f(a)\\}\) is closed in \(\mathbb{R}^{2}\).

Let \(f\) and \(g\) be continuous mappings from a normed linear space \(X\) into a second normed linear space \(Y\). Prove that \(\\{x: f(x)=\) \(g(x)\\}\) is closed in \(X\).

Show that \(\left\\{\left(a, \frac{1}{a}\right): a>0\right\\}\) is closed in \(\mathbb{R}^{2}\).

Let \(f\) be a continuous mapping from a normed linear space \(X\) into \(\mathbb{R}\). Then the following sets are open: $$ \\{x: f(x)<\alpha\\}, \quad\\{x: f(x)>\beta\\}, \quad\\{x: \alpha

Show that in \(\mathbb{R}^{3}\), (i) the set \(\\{(s, t, u): s, t, u>0\\}\) is open (ii) the set \(\\{(s, t, u): s=t=u\\}\) is not open.

Show that if \(G\) is open in the normed linear space \(X\) and \(A\) is a set with \(A \cap G\) empty, then \(\bar{A} \cap G\) is also empty.

Let \(A\) be a subset of the normed linear space \(X\) satisfying \(\lambda a \in A\) whenever \(a \in A\) and \(\lambda \geq 0\). Show that \(A\) is closed if and only if \(A \cap\\{x:\|x\| \leq 1\\}\) is closed.

For any set \(A\) in a normed linear space \(X\), show that \(\operatorname{dist}(x, A)=\) \(\operatorname{dist}(x, \bar{A})\) for all \(x\) in \(X\).

Show that \(\\{(s, 0): s \in \mathbb{R}\\}\) is closed in \(\mathbb{R}^{2}\) with any of the norms \(\|\cdot\|_{1},\|\cdot\|_{2}\) or \(\|\cdot\|_{\infty}\)

Define the distance between two sets \(A\) and \(B\) in a normed linear space \(X\) as $$ d(A, B)=\inf \\{\|a-b\|: a \in A, b \in B\\} $$ Is it possible to have two closed sets \(A\) and \(B\) in \(X\) such that \(d(A, B)=0\) and \(A \cap B\) empty? It might help to consider \(X=\mathbb{R}^{2}\), $$ A=\\{(s, 0): s \in \mathbb{R}\\} \quad \text { and } \quad B=\left\\{\left(s, e^{-s^{2}}\right): s \in \mathbb{R}\right\\} $$

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 \(B\).

Let \(\mathrm{E}\) be a linear subspace of the normed linear space \(X\). Suppose that for some \(\delta>0, E\) contains all \(x\) in \(X\) with \(\|x\|<\delta\). Show that \(E=X\). Deduce that if \(E \neq X\), then \(E\) has no interior points.

Give an example of a set \(A\) in \(\mathbb{R}^{2}\) such that \(\bar{A}=\mathbb{R}^{2}\) and int \(A\) is empty.

Let \(\left(X,\|\cdot\|_{X}\right)\) and \(\left(Y,\|\cdot\|_{Y}\right)\) be normed linear spaces. Set \(Z=\) \(\left(X \times Y,\|\cdot\|_{Z}\right)\), where as usual \(X \times Y\) is the set of ordered pairs \((x, y)\), where \(x\) lies in \(X\) and \(y\) in \(Y\). The norm is defined by $$ \|z\|_{z}=\|(x, y)\|_{z}=\max \left\\{\|x\|_{X},\|y\|_{Y}\right\\} $$ For sets \(A \in X\) and \(B \in Y\) show that (i) \(A \times B\) is open if and only if \(A\) and \(B\) are open (ii) \(A \times B\) is closed if and only if \(A\) and \(B\) are closed.

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.)