Infimum is a continuous function, compact set Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous map. Show that if $Y$ is compact then the function $g: X \rightarrow \mathbb{R}$ defined by $g(x) = \inf \{f(x,y): y \in Y\}$ is also continuous.
No clue here. Can you please help?
 A: Following the suggestion of Joe Johnson (I was going using intervals $(a,b)$, but this is simpler) (do you see why Joe's suggestion is enough? The sets $(a,\infty)$ and $(-\infty,a)$ form a subbasis for the topology of $\mathbb{R}$, and for a function to be continuous, it suffices to show that the inverse image of every set in a given subbasis is open).
Let $a\in\mathbb{R}$. What is $g^{-1}(-\infty,a)$? It consists of all $x\in X$ such that $g(x)\lt a$. If $g(x)\lt a$, there exists $y_x$ such that $f(x,y_x)\lt a$. Since $f$ is continuous, $f^{-1}(-\infty,a)$ is open, and $(x,y_x)\in f^{-1}(-\infty,a)$, so there exist open sets $U_x$ and $V_y$ of $X$ and $Y$, respectively, such that $U_x\times V_y\subseteq f^{-1}(-\infty,a)$. 
Now, suppose $x'\in U_x$. Then for all $y\in V_y$ (in particular, for $y_x$) you have $f(x',y_x)\in (-\infty,a)$. What does that tell you about $g(x')$? What does that tell you about $g^{-1}(-\infty,a)$?
Now try to do something along those lines with $g^{-1}(a,\infty)$. 
A: Old question, I know, but there’s a simple synthetic approach:
If $f: X × Y → ℝ$ is continuous, then so is the curried version $f^\mathrm{curr}\colon X → C(Y,ℝ)$, where $C(Y,ℝ)$ is the space of continuous functions $Y → ℝ$ with the compact-open topology. Since $Y$ is compact, taking infima is continuous as a map $\inf \colon C(Y,ℝ) → ℝ, f ↦ \inf f$. Thus $\inf ∘ f^\mathrm{curr} \colon X → ℝ$ is continuous as well, which is exactly the function in question.
A: Perhaps, we can suppose  that  $X$   and $Y$ are metric spaces (in particular   subset of real line $\mathbb R$)  and that $f(x,y)$    is real-valued function   in $x ,y$.   For a simple model we can   take $X=\mathbb R$  and $Y=[a,b]$ closed interval  in $\mathbb R$.
Set  $I(x) = \inf_{y\in Y} f(x,y)$.   Fix a point  $x_0$ in    $X$ .
Let us prove that   $I(x)$    is continuous at $x_0$.
Consider a sequence $x_n$   in $X$   such that   $x_n$    converges to  $x_0$.
Let us prove that    $I(x_n)$ converges to $I(x_0)$.
Since $Y$ is compact   there is a  sequence $(y_n)$    and  $y_0$ in $Y$   such that   $I(x_n) = f(x_n,y_n)$  and   $a_0:= I(x_0)= f(x_0,y_0)$.   Since $Y$ is compact there is a subsequence of $y_n$   which converges to $y^0$  in $Y$.  By abusing notation we can suppose that  $(x_n,y_n)$ converges to  $(x_0, y^0)$.  By continuity of $f$ in $x ,y$    we conclude that  $f(x_n,y_n)$  converges to   $f(x_0, y^0):=b_0$.  By definition of $I(x_n)$,        $f(x_n,y_n) \leq f(x_n,y_0)$  and therefore     since   $(x_n,y_n)$   converges to $(x_0, y^0)$   and    $(x_n,y_0)$    converges to $(x_0, y_0)$, by continuity of $f$,    we find
$b_0=f(x_0,y^0)\leq a_0$.  On the other hand,  by definition   $a_0\leq b_0$. Hence $a_0=b_0$.
This shows that the set of  cluster points   (or accumulation points) of sequence $I(x_n)$  is a  single point $a_0$  and therefore   $I(x_n)$  converges to  $a_0=I(x_0)$.
Therefore  $I$  is continuous in $x_0$ .
If necessary, we can give more details.
You can also see Mathematical Analyis I/II. by Vladimir A. Zorich. 
Can I get a text book reference for the proof that $g(x) = \inf_{y\in Y} f(x,y)$ is continuous in $x$ given compactness of $Y$ and continuity of $f$ in $x ,y$? - ResearchGate. Available from: https://www.researchgate.net/post/Can_I_get_a_text_book_reference_for_the_proof_that_gx_inf_yin_Y_fx_y_is_continuous_in_x_given_compactness_of_Y_and_continuity_of_f_in_x_y [accessed Jul 29, 2015].
