Proper map to locally compact Hausdorrf space is closed map via one point compactification.

Lemma. Let $f:X\to Y$$f:X\to Y$ be a continuous function from a compact space to Hausdorrf space. Then $f$$f$ is a closed map.

Proof. Let $F\subseteq X$$F\subseteq X$ be a closed set, hence it is compact, hence $f(F)$$f(F)$ is compact in $Y$$Y$, but $Y$$Y$ is Hausdorff space, hence $f(F)$$f(F)$ is closed. $◻$$\square$

Lemma. Let $f:X\to Y$$f:X\to Y$ be a continuous proper map,$Y$$Y$ is locally compact Hausdorrf space,then $f^{\ast }:X^{\ast }\to Y^{\ast }$$f^*:X^*\to Y^*$ is continuous.

Proof. Recall that the one-point compactification $X^{\ast }=X\cup {\mathrm{\infty }}_X$$X^*=X\cup{\infty_X}$ has the topology in which

1. Every open set $U\subseteq X$$U\subseteq X$ remains open in $X^{\ast }$$X^*$.

2. A neighborhood of ${\mathrm{\infty }}_X$$\infty_X$ is any set of the form $X^{\ast }\setminus K$$X^*\setminus K$, where $K\subseteq X$$K\subseteq X$ is compact and closed in $X$$X$.

Similarly for $Y^{\ast }=Y\cup {\mathrm{\infty }}_Y$$Y^*=Y\cup{\infty_Y}$. We define $\begin{array}{c}{f}^{\ast }:X^{\ast }⟶Y^{\ast },f^{\ast }(x)=\{\begin{array}{ll}f(x),& x\in X,\\ {\mathrm{\infty }}_Y,& x={\mathrm{\infty }}_X.\end{array}\end{array}$$f^*:X^*\longrightarrow Y^*,\qquad f^*(x)= \begin{cases} f(x),&x\in X,\\ \infty_Y,&x=\infty_X. \end{cases}$

We check continuity of $f^{\ast }$$f^*$ at two kinds of open set:

1. For open set $U\subset X$$U\subset X$

Since $f^{\ast }{|}_X=f$$f^*|_X = f$ and $f$$f$ is continuous $X\to Y$$X\to Y$, for any open $V\subseteq Y^{\ast }$$V\subseteq Y^*$ with $f^{\ast }(x)=f(x)\in V$$f^*(x)=f(x)\in V$ there is an open $U\subseteq X$$U\subseteq X$ with $x\in U$$x\in U$ and $f(U)\subseteq V$$f(U)\subseteq V$. But $U$$U$ is also open in $X^{\ast }$$X^*$, and $f^{\ast }(U)=f(U)\subseteq V$$f^*(U)=f(U)\subseteq V$. Thus $f^{\ast }$$f^*$ is continuous at each $x\in X$$x\in X$.

2. For open set contain ${\mathrm{\infty }}_X$$\infty_X$

Let $V\subseteq Y^{\ast }$$V\subseteq Y^*$ be any open neighborhood of ${\mathrm{\infty }}_Y=f^{\ast }({\mathrm{\infty }}_X)$$\infty_Y=f^*(\infty_X)$. By the topology of the one-point compactification, its complement $K=Y^{\ast }\setminus V$$K = Y^*\setminus V$ is a compact and closed subset of $Y$$Y$. Because $f$$f$ is proper, the inverse image $f^{-1}(K)\subseteq X$$f^{-1}(K)\subseteq X$ is compact and closed. Hence $W=X^{\ast }\setminus f^{-1}(K)={\mathrm{\infty }}_X\cup (X\setminus f^{-1}(K))$$W = X^*\setminus f^{-1}(K) = {\infty_X}\cup\bigl(X\setminus f^{-1}(K)\bigr)$ is an open neighborhood of ${\mathrm{\infty }}_X$$\infty_X$ in $X^{\ast }$$X^*$.

Having checked continuity at all points of $X^{\ast }$$X^*$, we conclude that $f^{\ast }:X^{\ast }⟶Y^{\ast }$$f^*:X^*\longrightarrow Y^*$ is indeed continuous. $◻$$\square$

Proposition. Let $f：X\to Y$$f：X\to Y$ be a proper map, where $Y$$Y$ is locally compcat Hausdorff space, then $f$$f$ is a closed map.

Proof. Apply the one point compactification we get $f^{\ast }:X^{\ast }\to Y^{\ast }$$f^*:X^*\to Y^*$ is a continous function from a compact space to a Hausdorrf space, hence $f^{\ast }$$f^*$ is closed.

Remark. $Y^{\ast }$$Y^*$ is Hausdorrf iff $Y$$Y$ is locally compact Hausdorrf.

Recall that the one-point compactification $X^{\ast }=X\cup {\mathrm{\infty }}_X$$X^*=X\cup{\infty_X}$ has the topology in which

1. Every open set $U\subseteq X$$U\subseteq X$ remains open in $X^{\ast }$$X^*$.

2. A neighborhood of ${\mathrm{\infty }}_X$$\infty_X$ is any set of the form $X^{\ast }\setminus K$$X^*\setminus K$, where $K\subseteq X$$K\subseteq X$ is compact and closed in $X$$X$.

Hence the closed set in $X^{\ast }$$X^*$ are $F\cup \{{\mathrm{\infty }}_X\}$$F\cup\{\infty_X\}$ where $F$$F$ is closed in $X$$X$ and $K$$K$, $K$$K$ is compact and closed.

Notice that if $F\subseteq X$$F\subseteq X$ is closed in $X$$X$, then $F\cup \{{\mathrm{\infty }}_X\}$$F\cup\{\infty_X\}$ is closed. Hence we have $f^{\ast }(F\cup \{{\mathrm{\infty }}_X\})=f(F)\cup {\mathrm{\infty }}_Y$$f^*(F\cup\{\infty_X\})=f(F)\cup\infty_Y$ is closed.

Hence $f(F)$$f(F)$ is closed in $Y$$Y$, $f$$f$ is a closed map. $◻$$\square$

Application. Define $f_a(x):=d_2(a,x):{\mathbb{R}}^n\to [0,\mathrm{\infty })$$f_a(x):=d_2(a,x):\mathbb R^n\to [0,\infty)$. Let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$$\Omega\subseteq \mathbb R^n$ be a non empty closed subset, then $f_a{|}_{\mathrm{\Omega }}$$f_a|_{\Omega}$ attains its minimum value.

Proof. It is easy to see that $f_a(x)$$f_a(x)$ is proper. Take any closed and bounded subset $U\subset [0,b]$$U\subset[0,b]$ then $f_a^{-1}(U)\subset f_a^{-1}([0,b])$$f_a^{-1}(U)\subset f_a^{-1}([0,b])$ is bounded and closed, hence compact. Notice that $[0,\mathrm{\infty })$$[0,\infty)$ is locally compact Hausdorrf, hence $f_a$$f_a$ is a closed map. Thus $f_a(\mathrm{\Omega })$$f_a(\Omega)$ is closed. $f_a(\mathrm{\Omega })$$f_a(\Omega)$ has lower bound hence $m=inf{f}_a(\mathrm{\Omega })$$m=\inf f_a(\Omega)$ exists. We need to prove that closed set contain its $inf$$\inf$.

By definition of $inf$$\inf$, $\mathrm{\forall }s\in f_a(\mathrm{\Omega }),m\le s.\mathrm{\forall }\epsilon >0,\mathrm{\exists }s\in f_a(\mathrm{\Omega }),s<m+\epsilon$$\forall\,s\in f_a(\Omega),\;m\le s. \forall\,\varepsilon>0,\;\exists\,s\in f_a(\Omega), s<m+\varepsilon$. Hence we could take $m<s_n<m+\frac{1}{n}$$m<s_n<m+\frac1n$, it is a sequence convergence to $m$$m$. Then it becomes a direct result of closed set contain the limit of the sequence on it. $◻$$\square$