1

It is easy to check that for any \((a_\alpha)_{\alpha\in J} \ne \mathbf{0}\) and \((b_\alpha)_{\alpha\in J}\) the maps \((x_\alpha)\mapsto (a_\alpha x_\alpha)\) and \((x_\alpha)\mapsto (x_\alpha + b_\alpha)\) are continuous in the box topology of \(\mathbb{R}^J\) (just take their pullbacks coordinate-wise). Therefore \((x_\alpha)\mapsto (a_\alpha x_\alpha + b_\alpha)\) is also continuous, and as its inverse is of similar form, a homeomorphism.

Therefore without loss of generality we only need to separate \(\mathbf{0}\) from the complement of \(\prod_{\alpha\in J} (-1, 1)\) using a \([0,1]\) valued continuous function. Consider \(\overline{\rho}\) the uniform metric in \(R_J\), that is \[\overline{\rho}(\mathbf{y},\mathbf{z}) = \sup_{\alpha\in J} \min(1,\left\lvert y_\alpha - z_\alpha\right\rvert).\] \(\overline{\rho}(\mathbf{0}, -)\) is a continuous (in the uniform topology) function which sends \(\mathbf{0}\) to \(0\) and the complement of \(\prod_J (-1,1)\) to \(1\). Now since the box topology is finer than the uniform topology, \(\overline{\rho}(\mathbf{0}, -)\) is also continuous in the box topology and we are done.

2

Let \(X\) be a regular Lindelöf space that is locally metrizable. By Urysohn metrization theorem we show that \(X\) has countable basis.

For each \(x\in X\) let \(U_x\) be a neighbourhood that is metrizable. As \(X\) is regular we have \(x \in V_x\subseteq \overline{V_x}\subseteq U_x\) where \(V_x\) is a smaller open neighbourhood. Since \(\overline{V_x}\) is a closed subspace, so it is also Lindelöf and as metrizability is hereditary \(\overline{V_x}\) has countable basis, and so does \(V_x\). Let \(\mathcal{B}_x\) be a countable basis for \(V_x\). As \(X\) is Lindelöf and \(\left\{V_x\right\}_{x\in X}\) is an open cover, we have a countable subcover \(V_{x_1}, V_{x_2},\dots\). To see that \(\mathcal{B} = \mathcal{B}_{x_1}\cup \mathcal{B}_{x_2}\cup\cdots\) is a base of \(X\), let \(U\subseteq X\) be open. For each \(x_i\), \(U\cap V_{x_i}\) is open in \(V_{x_i}\) and there is some subset \(\mathcal{C}_{x_i}\subseteq \mathcal{B}_{x_i}\) such that \(U\cap V_{x_i} = \bigcup \mathcal{C}_{x_i}\). Therefore \(U = \bigcup_i (U\cap V_{x_i}) = \bigcup_i \bigcup \mathcal{C}_{x_i}\) which is a union of sets from \(\mathcal{B}\). Note that \(\mathcal{B}\) is a countable union of countable sets and is hence countable.

3

Let \((X,d)\) be a complete metric space, let \(A_1\supset A_2\supset \dots\) be non-empty closed sets, let \(d_i = \sup_{x,y\in A_i} d(x,y)\) be the diameter of \(A_i\), and assume that \(\lim_{i\to\infty} d_i = 0\). Choose a sequence \(\mathbf{x} = (x_1,x_2,\dots) \in \prod_{i=1}^\infty A_i\). For every \(\varepsilon>0\) there is some \(N\) such that \(d_i < \varepsilon\) for all \(i\geq N\). Now as all \(x_N, x_{N+1},\dots \in A_N\) whose diameter is \(d_N < \varepsilon\), it follows that \(d(x_i,x_j) \leq d_N < \varepsilon\) for all \(i,j>N\) and hence \(\mathbf{x}\) is Cauchy. As a result \(x = \lim_{i\to\infty} x_i\) exists. For each \(i\), \(x\) is a limit of some sequence in \(A_i\), so \(x\in \bigcap_{i=1}^\infty A_i\) as each \(A_i\) is closed.

Conversely suppose every nested sequence of nonempty closed sets with diameter tending to 0 has a nonempty intersection, we want to show that \(X\) is complete. Let \((x_i)\) is be a Cauchy sequence and we want it to converge in \(X\). For each \(i\) let \(r_i = \sup_{j>i} d(x_i,x_j)\) and \(d_i = 2r_i\), note that diameter of \(\left\{x_i,x_{i+1},\dots\right\}\) is at most \(d_i\). As the sequence is Cauchy \(d_i \to 0\). Consider the nested sequence of non-empty closed balls \[\overline{B(x_1,r_1)}\supset \overline{B(x_2,r_2)}\supset\dots \] with diameter converging to \(0\). By hypothesis it has a nonempty intersection, let \(x\in \bigcap_{i=1}^\infty \overline{B(x_i,r_i)}\). We show \(x_i\to x\). Given \(\varepsilon>0\) as \((x_i)\) is Cauchy there is \(N\) such that for all \(j,k\geq N\), \(d(x_j, x_k) < \varepsilon\). In particular \(d(x_N, x_j) < \varepsilon\) for all \(j\geq N\) which means \(r_N < \varepsilon\). As \(x\in\overline{B(x_N,r_N)}\), \(d(x_N, x)\leq r_N\) and by triangle inequality for all \(j\geq N\), \(d(x,x_j) < 2\varepsilon\). This shows convergence.

4

(a)

Let \(\mathscr{F}\) be finite subset of \(\mathscr{C}(X,Y)\).

Induct on size of \(\mathscr{F}\). Case empty is trivial, case \(\left\lvert \mathscr{F}\right\rvert= 1\) is trivial too as its member is continuous. For the inductive case suppose \(\mathscr{F} = \mathscr{F}' \cup \left\{f\right\}\) and \(\mathscr{F}'\) is an equicontinuous family. Let \(x\in X\) and \(\varepsilon>0\), by induction hypothesis there is a neighbourhood \(U\) of \(x\) such that for all \(f'\in\mathscr{F}'\), \(f'(U)\subseteq B_\varepsilon(f'(x))\). As \(f\) is continuous, there is a neighbourhood \(V\) of \(x\) such that \(f(V)\subseteq B_\varepsilon(f(x))\). Now \(U\cap V\) is a neighbourhood of \(x\) satisfying the equicontinuity criterion.

(b)

Suppose \(f_n\) converges uniformly to \(f\), so \(\sup_{x\in X} d(f_n(x), f(x)) \to 0\). Let \(x\in X\) and \(\varepsilon>0\), As \(f\) is a uniform limit of continuous functions \(f\) is continuous, there is some neighbourhood \(U\) of \(x\) such that \(f(U)\subseteq B_{\varepsilon/3}(f(x))\). By the uniform convergence there is also some \(N\) such that for all \(n\geq N\), \(\overline{\rho}(f_n, f) < \varepsilon/3\). Now for each \(n\geq N, y\in U\), \[d(f_n(y), f_n(x)) \leq d(f_n(y), f(y)) + d(f(y), f(x)) + d(f(x), f_n(x)) < \varepsilon\] Also, \(\left\{f_1,\dots,f_{N-1}\right\}\) is an equicontinuous family by part (a) so let \(V\) be a neighbourhood witnessing equicontinuity of this set given \(\varepsilon\) and \(x\). Now for each \(n\), we have \(f_n(U\cap V)\subseteq B_\varepsilon(f_n(x))\) and we are done.

(c)

Fix \(x\in\mathbb{R}, \varepsilon>0\) and let \(U\) such that \(x\in U\in\mathscr{U}\). By hypothesis we know for each \(f\in\mathscr{F}\), \(\left\lvert f'(x)\right\rvert \leq M_U\). Choose \(\delta>0\) such that \(\delta< \frac{\varepsilon}{M_U}\) and \(B_\delta(x)\subseteq U\). It follow from mean value theorem that \(f(B_\delta(x)) \subseteq B_\varepsilon(f(x))\).

5

Suppose \(Y\) is Hausdorff, and let \(f\ne g\) in \(\mathscr{C}(X,Y)\). Let \(x\in X\) such that \(f(x)\ne g(x)\) and \(U,V\) such that \(f(x)\in U\) and \(g(x)\in V\), then \(S(\left\{x\right\},U)\) and \(S(\left\{x\right\},V)\) separate \(f\) and \(g\).

Suppose \(Y\) is regular, let \(f\in \mathscr{C}(X,Y)\) and \(S(C,V)\) a neighbourhood of \(f\). As \(f(C)\subseteq V\), \(\left\{f^{-1}(v)\right\}_{v\in V}\) is an open cover of \(C\) let \(f^{-1}(v_1),\dots,f^{-1}(v_n)\) be a finite subcover of \(C\). As \(Y\) is regular for each \(v_i\) let \(U_i\) be a neighbourhood such that \(v_i\in U_i \subseteq \overline{U_i} \subseteq V\). Set \(U = \bigcup_{i=1}^n U_i\) and \(\overline{U}\subseteq V\), we can see that \(f(C)\subseteq U\) so \(f\in S(C,U)\). Now we show \(\overline{S(C,U)}\subseteq S(C,V)\). If \(g\in\overline{S(C,U)}\), any neighbourhood of \(g\) meets \(S(C,U)\). Suppose for the contrary that \(g(C)\nsubseteq V\), so let \(x\in C\) such that \(g(x)\notin V\). In particular \(g(x)\notin\overline{U}\). Now \(S(\left\{x\right\}, Y\setminus\overline{U})\) is a neighbourhood of \(g\) that is disjoint from \(S(C,U)\), a contradiction. This result for sub-basic sets can be extended to all open sets which completes the proof of regularity.

6

To see \(f\) is continuous let \(x\in X, \varepsilon>0\).

By equicontinuity we have an open neighbourhood \(U\) of \(x\) such that for each \(n\), \(f_n(U)\subseteq B_{\varepsilon/2}(f_n(x))\). We want \(f(U)\subseteq B_{\varepsilon}(f(x))\). For \(u\in U\), use pointwise convergence to choose \(N\) such that for all \(n> N\), \(d(f_n(x), f(x)) < \varepsilon/4\) and \(d(f_n(u), f(u)) < \varepsilon/4\), then \[ d(f(u),f(x)) \leq d(f(u),f_n(u)) + d(f_n(u),f_n(x)) + d(f_n(x), f(x)) < \varepsilon\] so \(f\) is continuous.

To see compact convergence let \(K\subseteq X\) be compact and we show \(\sup_{x\in K} d(f_n(x), f(x)) \to 0\). Let \(\varepsilon> 0\), we want \(N\) such that \(n\geq N\) implies \(d(f_n(x), f(x)) < \varepsilon\) for all \(x\in K\). As \(f_n\) equicontiuous for each \(x\in K\) there is open neighbourhood \(U\) such that \(\sup_{y\in U} d(f_n(y),f_n(x)) < \varepsilon/6\). By covering \(K\) with finitely many such neighbourhoods \(U_1,\dots,U_m\) we have for each \(U_i\), \(\sup_{x,y\in U_i} d(f_n(x), f_n(y)) < \varepsilon/3\). Now use pointwise convergence to find \(N\) such that \(n\geq N\) implies \(d(f_n(x_i), f(x_i)) < \varepsilon/3\) for some chosen \(x_i\)’s where each \(x_i\in U_i\). And use continuity of \(f\) at each \(x_i\) such that \(N\) satisfies \(n\geq N\) implies \(d(f(x_i), f(x)) < \varepsilon/3\). Now for any \(x\in K\), \(n\geq N\), let \(x\) be covered by \(U_i\), then \[ d(f_n(x), f(x)) \leq d(f_n(x), f_n(x_i)) + d(f_n(x_i), f(x_i)) + d(f(x_i), f(x)) < \varepsilon. \]