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‘Proper’ is a heavily **overloaded **term, both in life and in mathematics. It may mean different stuff in different contexts. Thankfully mathematics is far less complicated that life and we can rigorously define **properness.**

**Proper Function:** A continuous function ( f : X \to Y ) between topological spaces is proper if the preimage of each compact subset of Y is compact.

**Proper Space: **A metric space (X, d) is proper if every closed ball ( B[x, r] = { y \in X | d(x, y) \leq r } ) in X is compact.

Intuition: **Proper Spaces **have much in common with Euclidean Spaces. Why? Heine Borel Theorem ensures that, closed balls are compact in Euclidean Space. Proper Spaces own this property as well.

There is a fancy description of Proper Spaces using Proper Functions.

Also see

Let (x_0 ) be an element of the metric space (X, d) and define ( d_0 : X \to [0,\infty) by d_0 (x) = d(x_0, x) ) . Then (X, d) is proper iff ( d_0 ) is proper.

We will rigorously prove this theorem. But first, lets draw some pictures.

**Driving Idea: **Which points in X map to the point ( a \in [0, \infty) ) ? They must be the points which are **a **distance away from ( x_0 ). Hence they are circles in X centered at ( x_0 ).

**(It is useful to try and write your own proofs first)**

Consider the compact set [0, a] (a is a finite number) in ( [0, \infty) ). This is a compact set (in ( [0, \infty) ).

Then its inverse image ( d_0^{-1} ([0, a) ) ) is compact in (X, d). (This is true because we have assumed (d_0 ) is a proper map, hence inverse images of compact sets will be compact).

But ( d_0^{-1} ([0, a) ) ) is a closed ball centered at ( x_0 ) of radius **a. **Thus we showed that all closed balls centered at ( x_0 ) are compact.

Next we will show that any other closed ball is compact. Consider a closed ball **B **of radius **r, centered at** ( x_1 \in X ). Suppose ( d(x_0, x_1) = t ) We take the ball **B’** of radius t + r centered at ( x_0 ).

If ( x \in B ) then ( d (x , x_1 ) \leq r ). Next we use the triangle inequality to compute:

$$ d(x_0, x) \leq d(x_0, x_1) + d(x_1, x) \leq t + r $$

Hence **x **is in ball **B’ centered at ( x_0 ) **. Therefore B is contained in B’.

Suppose ( { U_{\alpha} }*{\alpha \in \Lambda} ) is an arbitrary open cover of B (the closed ball centered at (x_1) ). Since B is closed (by assumption), therefore ( B^c ) is open and ( { { U*{\alpha} }_{\alpha \in \Lambda} , B^c } ) is an open cover for

Earlier we showed that all closed balls centered at ( x_0 ) are compact. B’ is one such balls. Hence it is compact. Hence ( { { U_{\alpha} }_{\alpha \in \Lambda} , B^c } ) has a finite subcover that covers B’. Since B is inside B’, this finite subcover also works for B’. If necessary, by removing ( B^c ), we get a finite subcover of B. Hence B is compact.

Suppose V is a compact subset of ( [0, \infty ) ). By Heine Borel Theorem, it is closed (contains all its limit points) and bounded. Since V is bounded, it has an upper bound. By Completeness axiom, it has a least upper bound **L**. Since it is closed, this least upper bound **L **is inside V.

( d_0^{-1} (L) ) is the circle of radius L centered at ( x_0 ). Every other point in the pre-image of V, is on a circle centered at ( x_0 ) of radius less than or equal to L. Hence the pre-image of V is bounded.

We will show that every infinite sequence has a convergent subsequence (sequential definition of compactness).

Suppose ( < x_n > ) is an infinite sequence in the pre image of V. Then ( d_0 ( x_n ) = a_n ) is an infinite sequence in V. As V is compact, it has a convergent subsequence (d_0 (x_{n_k}) = a_{n_k}) that converges to some ( a \in V ).

The preimage of **a** under (d_0 ), is the collection of all points from ( x_0 ) at **a** distance away. The distance from ( < x_{n_k} > ) to (B [x_0 , a] ) is 0 as ( d (x_{n_k} , x_0) ) approaches a as (n_k \to \infty ). If no point on ( \partial B ) (boundary of the ball) is a limit point of the sequence, then we can build an open cover using open balls ( U_x ) at each point on the boundary that completely misses the sequence (< x_{n_k} > ) and int (B).

Since ( B[x, a]) is compact, this open cover has a finite subcover. Let ( int (B), U_{x_1} , ... , U_{x_n} ) be the finite subcover. This leads to a contradiction as ( < x_{n_k} > ) cannot be closer to the boundary than the minimum of the distances of these finitely many open sets (each of which completely misses the sequence).

Suppose ( < x_k> ) is Cauchy in a proper space (X, d). That is ( \forall \epsilon > 0 \exists N \in \mathbb{N} \ni \forall m, n > N, d(x_m, x_n) < \epsilon ).

This implies ( | d(x_m, x_0) - d(x_n, x_0) | < d(x_m, x_n) < \epsilon ). Therefore ( |d_0 (x_m) - d_0 (x_n) | < \epsilon ) in ( [0, \infty) ).

Hence ( d_0 ( x_k) ) is Cauchy in ( [0, \infty) ). Since ( [0, \infty) ) is complete, therefore this converges to some nonnegative number **a.**

Finally using arguments similar to the last part of the previous proof, we are done.

**Theorem: **Show that a continuous function ( f : X \to Y ) between proper metric spaces is proper iff the obvious extension ( f : X^* \to Y^* ) between one-point compactification spaces is continuous.

**Proof: **

**Recall One Point Compactification: **

Put ( {\displaystyle X^{*}=X\cup {\infty }} ) , and topologize ( {\displaystyle X^{*}} ) by taking as open sets all the open subsets U of X together with all sets ( {\displaystyle V=(X\setminus C)\cup {\infty }} ) where C is closed and compact in X. Here, ( {\displaystyle X\setminus C} ) denotes setminus. Note that ( {\displaystyle V} ) is an open neighborhood of ( {\displaystyle {\infty }} ), and thus, any open cover of ( {\displaystyle {\infty }} ) will contain all except a compact subset ( {\displaystyle C} ) of ( {\displaystyle X^{*}} ) , implying that ( {\displaystyle X^{*}} ) is compact.

**Proper --> Continuous**

Suppose ( f : X \to Y ) is proper. We will show that ( f^* ), the extension of f, is continuous. Toward that extent, we will show that inverse of any open set is open. Clearly if, ( V \subset X^* ) does not contain ( \infty_Y ) then its preimage is open in ( X^* ) as f is continuous.

If ( V \subset Y^* ) contains ( \infty_Y ) then( {\displaystyle V=(Y \setminus C)\cup {\infty_Y }} ) where C is closed and compact in X. Then

( \displaystyle { {f^*}^{-1} ( (Y \setminus C) \cup { \infty_Y } ) \ = { {f^*}^{-1} (Y) - {f^*}^{-1} ( C) } \cup {f^*}^{-1} { \infty_Y } \ = {X - f^{-1}(C)} \cup {\infty_X } } )

SInce f is proper ( f^{-1} (C) ) is compact and we have an open set of ( X^* ) as the preimage.

**Continuous --> Proper**

Suppose V is an open set in ( Y^* ) containing ( \infty_Y ). Since ( f^* ) is continuous, its preimage must be open.

( {\displaystyle V=(Y \setminus C)\cup {\infty_Y }} ) where C is closed and compact in X

( \displaystyle { {f^*}^{-1} ( (Y \setminus C) \cup { \infty_Y } ) \ = { {f^*}^{-1} (Y) - {f^*}^{-1} ( C) } \cup {f^*}^{-1} { \infty_Y } \ = {X - f^{-1}(C)} \cup {\infty_X } } )

This implies ( {X - f^{-1}(C)} \cup {\infty_X } ) is open in ( X^* ) containing ( \infty ). This implies ( f^{-1} (C) ) is compact (due to the characterization of open sets containing infinity). SInce for all compact subsets C of Y, the above argument is valid, therefore f is proper.

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