The continuity of infimum of a function

Question

Let $(X,d)$ be a connected metric space and $(Y,d')$ is a compact metric space. Let $f$ be a continuous function from $(X\times Y,\max(d,d'))$ into $\mathbb{R}$. Because $Y$ is compact we can define:

$$g(x)=\inf_{y\in Y}f(x,y)$$

for all $x\in X$. I need to prove that $g$ is continuous.

First I tried to prove that if $(x_n)_{n\ge 0}\rightarrow a$ is a convergent sequences in $X$, $(g(x_n))_{n\ge 0}$ converges to $g(a)$. I arrived at showing that $(g(x_n))_{n\ge 0}$ has only one adherent value, which is $g(a)$, but no more.

Then I tried to use the fact that for fixed $x$, the function $f(x,y)$ is uniformly continuous in $y$, but there was not so much progress.

In either attempts, I never used the connectedness of $X$, so I think that's the key, but I don't know how to implement this.

Answer 1

First note that the topology induced by the metric $\max(d,d')$ is just the product topology. That means we have continuous open projections $\pi_X: X\times Y\to X$ and $\pi_Y: X\times Y\to Y$.

Show that the preimage of an open subbase set $(a,\infty)$ under $g$ is open: This is the set $X-\{x\mid \exists y\in Y,f(x,y)\le a\}$. The complement can be expressed as $\pi_X(f^{-1}((\infty,a]))$. Since $(\infty,a]$ is closed, its preimage under $f$ is closed. Applying the projection $\pi_{\Bbb R}$, we get a closed set again. This is because the projection $X\times Y\to X$ is closed if $Y$ is compact.

For an open subbase set $(\infty,a)$, its preimage under $g$ is the set $\{x\mid \exists y\in Y,f(x,y)<a\}$, which is $\pi_X(f^{-1}((\infty,a)))$.

Connectedness of $X$ is not needed.

See also Continuity of max (moving the domain and the function), for a similar result.