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In the metric space $(Z,d)$, let $A(z_0,\varepsilon)$ denote the closed ball $\left\lbrace z\mid d(z,z_0)\leq\varepsilon\right\rbrace$. Now let $X$ be an arbitrary space, let $Y$ be a metric space and let $f:X \times Y \rightarrow Z$ be continuous in each variable separately. Let $\varepsilon>0$ and $y_0 \in Y$ be kept fixed, and for each $x \in X$ define $d(x)=\sup\left\lbrace r\mid f[x,B(y_0,r)]\subset A[f(x,y_0),\varepsilon]\right\rbrace$. Prove:

  • $x \rightarrow d(x)$ is an upper semicontinuous map of $X$ into $E^1$.
  • If $X$ is a Baire space, then there exists an $x_0 \in X$ such that $d\left\lbrace F(x,y),F(x_0,y_0)\right\rbrace\leq 2\varepsilon$ on some nbd $U(x_0) \times V(y_0)$

Def: An $f: X \rightarrow E^1$ is upper semicontinuous if for each real b, $\left\lbrace x\mid f(x)<b\right\rbrace$ is open.

Definition of Baire space: Given any countable collection $\left\lbrace A_n \right\rbrace$ of closed sets of X each of which has empty interior in X, their union $\cup A_n$ also has empty interior in X.

I would like you tell me some ideas to do the exercise. Thanks for your help

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Denote the metric of the space $Y$ by $d_Y$.

1) Let $x\in X$, $b\in\Bbb R$ and $d(x)<b$. So there exists a point $y\in Y$ such that $d_Y(y,y_0)<b$ but $d(f(x,y),f(x,y_0))>\varepsilon$. Put $\varepsilon’=d(f(x,y),f(x,y_0))- \varepsilon >0$. The separate continuity of the function $f$ implies that there exists a neighborhood $O(x)$ of the point $x$ such that

$$d(f(x’,y_0), f(x,y_0))<\varepsilon’/2\mbox{ for each }x’\in O(x).$$

and a neighborhood $O'(x)$ of the point $x$ such that

$$d(f(x’,y), f(x,y))<\varepsilon’/2\mbox{ for each }x’\in O(x).$$

Let $x’\in O(x)\cap O’(x)$ be an arbitrary point. Then by the triangle inequality

$$ d(f(x,y),f(x,y_0))\le d(f(x,y),f(x’,y))+ d(f(x’,y),f(x’,y_0))+ d(f(x’,y_0),f(x,y_0)).$$

So

$$ d(f(x’,y),f(x’,y_0))\ge d(f(x,y),f(x,y_0))- d(f(x,y),f(x’,y))- d(f(x’,y_0),f(x,y_0))>\varepsilon+\varepsilon’- \varepsilon’/2-\varepsilon’/2=\varepsilon,$$

thus $d(x’)<b$.

2) Assume that $X$ is a Baire space and $\varepsilon>0$ be an arbitrary real numberFor each natural $n$ put

$$X_n=\{x\in X: d(f(x,y), f(x,y_0)\le\varepsilon\mbox{ for each }y\in Y\mbox{ such that }d_Y(y,y_0)\le 1/n\}.$$

We claim that each $X_n$ is a closed subset of the space $X$. Indeed, assume the converse, there exists a point $x\in\overline{X_n}$ and a point $y\in Y$ such that $d_Y(y,y_0)\le 1/n$ but $$d(f(x,y), f(x,y_0))-\varepsilon=\varepsilon’>0.$$ The separate continuity of the function $f$ implies that there exist a neighborhood $O(x)$ of the point $x$ such that

$$d(f(x’,y_0), f(x,y_0))<\varepsilon’/2\mbox{ for each }x’\in O(x)$$

and a neighborhood $O'(x)$ of the point $x$ such that

$$d(f(x’,y), f(x,y))<\varepsilon’/2\mbox{ for each }x’\in O(x).$$

Since $x\in\overline{X_n}$, there exist a point $x’\in O(x)\cap O'(x)\cap X_n$. Then by the triangle inequality

$$\varepsilon+\varepsilon’=d(f(x,y), f(x,y_0))\le$$ $$ d(f(x,y), f(x',y))+d(f(x',y), f(x',y_0))+d(f(x',y_0), f(x,y_0))<$$ $$\varepsilon’/2+\varepsilon+\varepsilon’/2=\varepsilon+\varepsilon’,$$

a contradiction.

The separate continuity of the function $f$ implies that $X=\bigcup X_n$. Since the space $X$ is Baire, there exist a natural number $n$ such that the set $X_n$ has non-empty interior. Pick any point $x_0$ contained in this interior and an arbitrary its neighborhood $U_0(x_0)\subset X_n$. The separate continuity of the function $f$ implies that there exists a neighborhood $U_1(x_0)$ of the point $x_0$ such that

$$d(f(x,y_0), f(x_0,y_0))\le\varepsilon\mbox{ for each }x\in U_1(x_0).$$

Put $$U(x_0)=U_0(x_0)\cap U_1(x_0)\mbox{ and }V(y_0)=\{y\in Y: d(y,y_0)\le 1/n\}.$$

Let $(x,y)\in U(x_0)\times V(y_0)$ be an arbitrary point. Then

$$ d(f(x,y), f(x_0,y_0))\le d(f(x,y), f(x,y_0))+d(f(x,y_0), f(x_0,y_0)) \le \varepsilon+\varepsilon=2\varepsilon.$$

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