Image of a compact set through a continuous open mapping.

Question

Suppose $X$ and $Y$ are metric spaces, $f:X\rightarrow Y$ is a continuous open mapping and $F\subseteq X$ is a compact set. Show for each $x\in F$ and $r>0$ there exists $w=w(r)>0$ such that $$B^Y_{w(r)}(f(x))\subseteq f\left(B^X_r(x)\right).$$ I know $w(r,x)$ must necessarily exist due to the openness of the mapping. However I cannot show why it may be independent of $x$. I believe an application of the Lebesgue Covering Lemma may help but I am not sure. Any help would be greatly appreciated.

Answer 1

Two ways:

1. Fix $r > 0$ and for each $x \in F$ let $w_x > 0$ be such that $f \left[ B^X \left( x, \frac{r}{2} \right) \right] \supseteq B^Y(f(x), 2w_x)$. Then $(x, f(x)) \in B^X \left( x, \frac{r}{2} \right) \times B^Y(f(x), w_x)$, so the family $\left\{ B^X \left( x, \frac{r}{2} \right) \times B^Y(f(x), w_x) : x \in F \right\}$ is an open cover of $\operatorname{gr} f \restriction F \subseteq X \times Y$. Since a graph of a continuous function on a compact set is compact, there is a finite subcover corresponding to some $x_1, \ldots, x_n \in F$. We will show that $w = \min \{ w_{x_1}, \ldots, w_{x_n} \}$ has the required property.

   So take any $x \in F$ and pick $i \in \{ 1, \ldots, n \}$ such that $(x, f(x)) \in B^X \left( x_i, \frac{r}{2} \right) \times B^Y(f(x_i), w_{x_i})$. Then $d^X(x, x_i) < \frac{r}{2}$ and $d^Y(f(x), f(x_i)) < w_{x_i}$, hence

   $B^Y(f(x), w) \subseteq B^Y(f(x), w_{x_i}) \subseteq B^Y(f(x_i), 2w_{x_i}) \subseteq f \left[ B^X \left( x_i, \frac{r}{2} \right) \right] \subseteq f \left[ B^X (x, r) \right],$

   as required.

2. Fix $r > 0$ and define $v : X \to (0, \infty)$ such that

   $$v(x) = \sup \left\{ w > 0 : (\exists r' < r) \, B^Y(f(x), w) \subseteq f \big[ B^X(x, r') \big] \right\}.$$

   First we will show that $v$ is lower semi-continuous. Fix $x \in X$ and $\varepsilon > 0$. We need to find $\delta > 0$ such that $v(y) \geqslant v(x) - 2\varepsilon$ for any $y \in X$ with $d^X(x, y) \leqslant \delta$. By definition of $v(x)$, there are $w \geqslant v(x) - \varepsilon$ and $r' < r$ such that $B^Y(f(x), w) \subseteq f \big[ B^X(x, r') \big]$. Take $0 < \delta < r-r'$ such that $d^Y(f(x), f(y)) \leqslant \varepsilon$ whenever $d^X(x, y) \leqslant \delta$. We claim that it satisfies the required condition. So fix any $y \in X$ and suppose $d^X(x, y) \leqslant \delta$. It is enough to show that $v(y) \geqslant w - \varepsilon$. Take $s = d^X(x, y) + r' < r$ so that $B^X(x, r') \subseteq B^X(y, s)$. Then $s$ witnesses that $w-\varepsilon$ belongs to the set in the definition of $v(y)$, because

   $$B^Y(f(y), w-\varepsilon) \subseteq B^Y(f(x), w) \subseteq f \big[ B^X(x, r') \big] \subseteq f \big[ B^X(y, s) \big].$$

   Hence $v(y) \geqslant w-\varepsilon$, as required.

   Now recall that a lower semi-continuous real function attains a minimum on a compact set. So there is $w > 0$ such that $v(x) > w$ for each $x \in F$. Then $w$ has the desired property, as for any $x \in X$ there is $w < w' \leqslant v(x)$ and $r' < r$ such that $B^Y(f(x), w') \subseteq f \big[ B^X(x, r') \big]$. In this setting

$$B^Y(f(x), w) \subseteq B^Y(f(x), w') \subseteq f \big[ B^X(x, r') \big] \subseteq f \big[ B^X(x, r) \big].$$