Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces; give \(X \times Y\) the corresponding square metric; let \(\mathcal{H}\) denote the collection of all nonempty closed, bounded subsets of \(X \times Y\), in the resulting Hausdorff metric. Consider the space \(\mathcal{C}(X, Y)\) in the uniform metric; let \(\operatorname{gr}: \mathcal{C}(X, Y) \to \mathcal{H}\) be the function that assigns, to each continuous function \(f: X \to Y\), its graph
\[ G_f = \ts{x \times f(x) \mid x \in X}. \]
Show that the map \(\operatorname{gr}\) is injective and uniformly continuous.
Let \(\mathcal{H}_0\) denote the image set of the map \(\operatorname{gr}\); let \(g: \mathcal{C}(X, Y) \to \mathcal{H}_0\) be the surjective map obtained from \(\operatorname{gr}\). Show that if \(f: X \to Y\) is uniformly continuous, then the map \(g^{-1}\) is continuous at the point \(G_f\).
Give an example where \(g^{-1}\) is not continuous at the point \(G_f\).
Theorem. If \(X\) is compact, then \(\operatorname{gr}: \mathcal{C}(X, Y) \to \mathcal{H}\) is an imbedding.
Solution
Write \(d_\square((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')\}\) and \(d_H\) for the Hausdorff metric.
Distinct functions have distinct graphs, so \(\operatorname{gr}\) is injective. If \(\|f-h\|_\infty<\varepsilon\), then for every \(x\), \((x,f(x))\) lies within \(\varepsilon\) of \((x,h(x))\), and conversely. Thus \[ d_H(G_f,G_h)\le \|f-h\|_\infty, \] so the graph map is \(1\)-Lipschitz.
Fix uniformly continuous \(f\) and \(\varepsilon>0\). Choose \(\delta<\varepsilon/2\) so that \(d_X(x,x')<\delta\) implies \(d_Y(f(x),f(x'))<\varepsilon/2\). If \(d_H(G_f,G_h)<\delta\), then for each \(x\) there is \(x'\) with \[ d_X(x,x')<\delta,\qquad d_Y(h(x),f(x'))<\delta. \] Hence \[ d_Y(h(x),f(x))<\delta+\varepsilon/2<\varepsilon. \] Taking the supremum gives \(\|h-f\|_\infty<\varepsilon\). Thus \(g^{-1}\) is continuous at \(G_f\).
Take \(X=\{1/k: k\in\mathbb Z_+\}\subset\mathbb R\) and \(Y=\mathbb R\), with the usual metrics. Every function on the discrete space \(X\) is continuous. Let \(f(1/k)=(-1)^k\), and for \(n\ge1\) let \(h_n(1/k)=(-1)^k\) for \(k\le n\) and \(h_n(1/k)=(-1)^{k+1}\) for \(k>n\). The graphs lie in \((0,1]\times[-1,1]\), so they are bounded, and they are closed because the functions are continuous.
For \(k>n\), the point \((1/k,h_n(1/k))=(1/k,f(1/(k+1)))\) of \(G_{h_n}\) lies within \(1/k-1/(k+1)<1/n^2\) of the point \((1/(k+1),f(1/(k+1)))\) of \(G_f\). Likewise \((1/k,f(1/k))=(1/k,h_n(1/(k+1)))\) lies within \(1/n^2\) of \(G_{h_n}\). The points with \(k\le n\) are common to both graphs. Hence \[ d_H(G_f,G_{h_n})\le 1/n^2\longrightarrow0. \] But \(|h_n(1/(n+1))-f(1/(n+1))|=2\), so \(\|h_n-f\|_\infty=2\) for every \(n\). Thus \(g^{-1}\) is not continuous at \(G_f\). Here \(f\) is not uniformly continuous, since \(|f(1/k)-f(1/(k+1))|=2\) while \(1/k-1/(k+1)\to0\).
- If \(X\) is compact, every continuous \(f:X\to Y\) is uniformly continuous. By (b), the inverse from the image \(\mathcal H_0\) to \(C(X,Y)\) is continuous at every graph. Together with (a), \(\operatorname{gr}\) is a homeomorphism onto its image, hence an embedding.