‘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

## Theorem:

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 \( x_0 \).

*Proof:*

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

#### —> Suppose \( d_0 \) is a proper map. Then we will show that (X, d) is a proper space (that is all closed balls are compact).

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 **B’. **

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 (X, d) is proper. Then we will show that \( d_0 \) is a proper map.

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).

## Proper Spaces share more properties with Euclidean Spaces. For example, every proper metric space is complete.

### Proof:

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.

## One Point Compactification

**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.