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.

Also see

College Mathematics Program at Cheenta

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.

Compact Set, Proper Spaces and Annulus

Euclidean Spaces have a very nice property. In ( \mathbb{R}^n ) (equipped with standard Euclidean metric), every closed and bounded set is a compact set. The converse is also true. Every compact set is closed and bounded). This property is known as Heine Borel Theorem.

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TIFR 2015 Problem 2 Solution -Image of continuous function


TIFR 2015 Problem 2 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:

Let (f: \mathbb{R} \to \mathbb{R} ) be a continuous function. Which of the following can not be the image of ((0,1]) under (f)?

A. {0}

B. ((0,1))

C. ([0,1))

D. ([0,1])

Discussion:

If f is the constant function constantly mapping to 0, which is continuous, then the image set is {0}.

Suppose that (f((0,1])=(0,1)) . Then (f((0,1))=(0,1)- {f(1)} ). Now since (f(1)\in (0,1) ) the set ( (0,1)- {f(1)} ) is not connected. But ((0,1)) is connected, and we know that continuous image of a connected set is connected. This gives a contradiction. So ((0,1)) can not be the image of ((0,1]) under f.

Define (f(x)=1-x).  Then (f((0,1])= [0,1)).

Define (f(x)=0) for (x\in [0,\frac{1}{2}] ) and (f(x)= 2(x-\frac{1}{2}) ) for (x\in [\frac{1}{2} ,1 ] ). (f) is continuous on  ((0,\frac{1}{2}] ) and ( [\frac{1}{2} ,1 ] ) and (f) agrees on the common points, by pasting lemma (f) is continuous on ( [0,1] ) . And image of ((0,1] ) is ([0,1]).

TIFR 2015 Problem 2 Solution is concluded.

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TIFR 2014 Problem 28 Solution - Continuous Functions from Discrete Space


TIFR 2014 Problem 28 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:

Let $X$ be a topological space such that every function $f: X \to \mathbb{R}$ is continuous. Then

A. $X$ has the discrete topology.

B. $X$ has the indiscrete topology.

C. $X$ is compact.

D. $X$ is not connected.

Discussion:

We know that if $Y$ is a discrete space then any function $g: Y \to Z $ is continuous.

Option A asks whether the converse to this is true in the case that $Z= \mathbb{R}$.

To prove/disprove whether $X$ has the discrete topology or not it is enough to prove whether every singleton set is open or not.

If we can show that for every $x\in X$ there exists a function $f_x :X \to \mathbb{R}$ such that $f_x^{-1} (-1,1) = {x}$ then we are done. Because we are given that $f_x$ if exists must be continuous, and since $(-1,1)$ is open in $\mathbb{R}$ we will have the inverse image of it open in $X$, so $x $ will be open in $X$.

Now, this target is easy to handle. We define for each $x\in X$

$f_x (x) = 0$ and $f_x (y) =2$ for $y \neq x $.

This $f_x$ satisfies our desired property. So $X$ is discrete.

Taking $X= \mathbb{Z}$ (for example) shows that $X$  does not need to be indiscrete nor does it have to be compact.

Taking $X= {0}$ shows that $X$ may be connected. Of course if $X$ has cardinality more than 1, it is not connected.

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TIFR 2014 Problem 27 Solution -Homeomorphism of product space with mother space


TIFR 2014 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 Topology by James.R. Munkres. This book is very useful for the preparation of TIFR Entrance.

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Problem:

(X) is a topological space with infinite cardinality which is homeomorphic to (X \times X). Then

A. (X) is not connected.

B. (X) is not compact.

C. (X) is not homeomorphic to a subset of (\mathbb{R})

D. none of the above

Discussion:

Let us take (X=\mathbb{Z}). The topology is the subspace topology obtained from (\mathbb{R}). Then, in fact, (X) has the discrete topology. This is because for every (n\in \mathbb{Z}), ((n-1,n+1)) is an open subset of (\mathbb{R}) and hence ((n-1,n+1)\cap \mathbb{Z}={n} ) is open in (\mathbb{Z}).

We have a bijection from (\mathbb{Z} \times \mathbb{Z}) to (\mathbb{Z}). The product of two discrete spaces is again a discrete space because we have all sets of the form ({n} \times {m} ={(n,m)} ) as open sets. Now, any map from a discrete space is continuous. Because the inverse image of any open set is a subset of the space and every subset is open in that space. So in fact, the bijection that we know of from (\mathbb{Z} \times \mathbb{Z}) to (\mathbb{Z}) is continuous in both ways. Therefore in this setup (X \times X) is homeomorphic to (X). And also (X=\mathbb{Z}) is a subset of (\mathbb{R}). So this provides us with a counterexample for option C.

Next, we will take (X=\mathbb{Z}) to disprove A and B together. But now we will consider a different topology for (\mathbb{Z}). We will consider the indiscrete topology. That is, the only sets which are open in (X) are (X) and the empty set. Note firstly, that since we only have finitely many open sets to start with, (X) is compact. (X) is also connected because the only non-empty set which is both open and closed is (X) itself. We use the same bijection that we know of between (\mathbb{Z}) and (\mathbb{Z}\times \mathbb{Z}). Note that if a map has its co-domain set as an indiscrete space then that map must be continuous. This is because the only open sets are empty and whole set, and their inverse images are empty and the whole domain set respectively which are always open. Also, note that if (X) has the indiscrete topology then so does (X \times X). Because the open sets in (X \times X) are (\phi \times \phi = \phi , \phi \times X = \phi , X \times \phi = \phi, X \times X ). So both the spaces in consideration have indiscrete topology. So the bijection that we have is continuous in both ways. This, therefore, gives the counterexample for A and B.

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TIFR 2014 Problem 15 Solution -Discrete space


TIFR 2014 Problem 15 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:

(X) is a metric space. (Y) is a closed subset of (X) such that the distance between any two points in (Y) is at most 1. Then

A. (Y) is compact.

B. any continuous function from (Y\to \mathbb{R}) is bounded.

C. (Y) is not an open subset of (X)

D. none of the above.

Discussion:

Let (X=) an infinite set for example (=\mathbb{R}) with the metric as discrete metric.

That is (d(x,y)=1) if (x\neq y) and (d(x,y)=0) if (x=y).

Then every set in (X) is open and every set is closed.

Now take (Y=X). Then (Y) can be covered by singleton sets. (Y=\cup {{a}|a\in Y}). Now each of the singleton sets is open in discrete metric space. Therefore, this is an open cover for (Y). Since (Y) is infinite, this cover has no finite subcover. So (Y) is not compact.

Given any (f:Y\to \mathbb{R}), for open set (U\in \mathbb{R}), (f^{-1}(U)\subset Y). Since Y is discrete, (f^{-1}(U))is open in (Y). So every function (f:Y\to \mathbb{R}) is a continuous function. In particular if we define (f(x)=x) then (f) is a continuous function. And (f) is not bounded.

Also, every subset of (X) is open. So (Y) is open.

Therefore, we are left with none of the above.

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Continuous Surjection ~ nonexistence | TIFR 2013 problem 26

Question:

True/False?

There exists a continuous surjective map from the complex plane onto the non-zero reals.

Hint:

Search for topological invariants.

Discussion:

Under a continuous function, connected set must go to connected set. The complex plane (\mathbb{C}) is connected.

It's image must be connected.

(\mathbb{R}-0) is not connected.

So the statement is False.

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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Non-homeomorphic (TIFR 2013 Math Solution problem 22)

TIFR 2013 Math Solution Discussion

Question:

The sets \([0,1)\) and \((0,1)\) are homeomorphic.

Hint:

Check some topological invariant.

TIFR 2013 Math Solution

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Discussion:

In \([0,1)\), \(0\) seems to be a special point, as compared to \((0,1)\) where every point has equal importance (or non-importance).

If \(f:X\to Y \) is a homeomorphism then for any point \(a\in X\),  \(X- \{ a \} \) and \(Y- \{ f(a) \} \) are homeomorphic with the homeomorphism function being restriction of \(f\) to \(X- \{a \} \).

If \(f: [0,1) \to (0,1) \) is a homeomorphism, then choosing to remove \(0\) from \([0,1)\) we get that \( (0,1) \) is homeomorphic to \( (0,1)- \{f(0) \} \).

Whatever be \(f(0) \), we see that \( (0,1)- \{f(0) \} \) is a disconnected set. Whereas, \((0,1)\) is connected. A continuous image of a connected set is always connected. Hence we are forced to conclude that there was no such \(f\) to begin with.

So the statement is False.