Proper Metric Spaces can be modeled by Rays!

‘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.

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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 ).

Proper Metric Spaces

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.

TIFR 2013 Problem 27 Solution -Bounded derivatives and Compactness


TIFR 2013 Problem 27 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:True/False?

If every differentiable function on a subset \(X\subset \mathbb{R}^n \) (i.e, restriction of a differentiable function on a neighbourhood of \(X\) ) is bounded, then \(X\) is compact.

Hint:

Recall that  \(X\subset \mathbb{R}^n \) is compact if and only if it is closed and bounded. Find suitable functions to show bounded and closed respectively.

Discussion:

Let \(f(x)=(||x||^2,||x||^2,...,||x||^2)\), then \(f\) is differentiable. Hence \(||x||\) must be bounded on \(X\), which is the same as saying \(X\) is bounded.

Note, \(||x||\) is the standard euclidean norm. And \(f\) is differentiable because each component of \(f\) is differentiable.

Now all we need is a function which will show that \(X\) is closed as well.

Let \(p\in \mathbb{R}^n \) be a point which is a limit point of \(X\). Suppose that \(p\notin X \). Then \(||x-p|| \ne 0 \) for any \(x\in X\). Hence the function \(g(x)=(\frac{1}{||x-p||},\frac{1}{||x-p||},...,\frac{1}{||x-p||} )  \) is a well-defined and differentiable function. Again, the differentiation part follows from the fact that it is differentiable in each component. And that it is differentiable in each component is easy to check. (Hint: Take the partial derivatives).

Now that we have \(g\), a differentiable function, we know that \(g\) must be bounded on \(X\). But as \(x \to p \) we see that \(||g(x)|| \to \infty \), i.e, g is not bounded. The only assumption we have made so far is that \(p\) is not in \(X\). This must be false, so \(p\in X\). So in fact, every limit point of \(X\) is in \(X\). This shows that \(X\) is closed.

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TIFR 2013 Problem 24 Solution -Non-existence of continuous function


TIFR 2013 Problem 24 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects. In general, TIFR entrance exam hits the floor during the month of December.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:True/False

There exists a continuous surjective function from \(S^1 \) onto \(\mathbb{R}\).

Hint:

Search for topological invariants.

Discussion:

We know that continuous image of a compact set is compact. \(S^1\) is a subset of \(\mathbb{R}^2\), and in \(\mathbb{R}^2\) a set is compact if and only if it is closed and bounded.

By definition, every element of \(S^1\) has unit modulus, so it is bounded.

Let's say \(z_n\to z\) as \(n\to \infty \). Where {\(z_n\)} is a sequence in \(S^1\). Since modulus is a continuous function, \(|z_n| \to |z| \), the sequence {\(|z_n|\)} is simply the constant sequence \(1,1,1,... \) hence \(|z|=1\).

What does above discussion mean? Well it means that if \(z\) is a limit point (or even a point of closure) of \(S^1\) then \(z\in S^1\).  Therefore, \(S^1\) is closed.

The immediate consequence is that the given statement is False. Because, \(\mathbb{R}\) is not compact. \(S^1\) is compact, and continuous image of a compact set has to be compact.

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TIFR 2013 problem 23 | Complete-Not Compact

Try this problem 23 from TIFR 2013 named - Complete not compact.

Question: TIFR 2013 problem 23

True/False?

Let \(X\) be complete metric space such that distance between any two points is less than 1. Then \(X\) is compact.

Hint:

What happens if you take discrete space?

Discussion:

Discrete metric space as we know it doesn't satisfy the distance < 1 condition. But we can make slight changes to serve our purpose.

In \(X\) define \(d(x,y)=\frac{1}{2}\) if \(x\ne y\). Otherwise, \(d(x,x)=0\).

\(d\) is indeed a metric, and it gives the same discrete topology on \(X\). Namely, every set is open because every singleton is open. And therefore every set is closed.

We want \(X\) to be complete. If \(x_n\) is a sequence in \(X\) which is Cauchy, then taking \(\epsilon=\frac{1}{4}\) in the definition of Cauchy sequence, we conclude that the sequence is eventually constant.

Since the tail of the sequence is constant, the sequence converges (to that constant).

This shows that \(X\) is indeed Complete.

We don't want \(X\) to be compact. Not all \(X\) will serve that purpose, for example a finite set is always compact. We take a particular \(X=\mathbb{R}\).

Since singleton sets are open, if we cover \(X\) by all singleton sets, then that cover has no finite subcover. Hence \(X\) is not compact.

Therefore the given statement is False.

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