An example: continuous map which is not homeomorphic

Reference: Topology by Hocking and Young. Dover Publications.

Let us present an example of a continuous mapping (one-to-one) which is not a homeomorphism:

Let S be the set of all non-negative real numbers with their usual metric topology, and let T be the unit circle in its metric topology. Consider $f: S \rightarrow T$. For each x in S, let $f(x) = (1, \frac{2\pi x^{2}}{(1+x^{2})})$, a point in polar coordinates on T.

It is easily shown that f is continuous and one-to-one.

But the set of all x in S such that $x<1$ is open in S while its image is not open in T. Hence, f is not interior. (meaning that: A transformation $f: S \rightarrow T$ of the space S into the space T is said to be interior if f is continuous and and if the image of every open subset of S is open in T) , and is not a homeomorphism (because of the following theorem: A necessary and sufficient condition that the one-to-one continuous map $f: S \rightarrow T$ of the space S onto the space S be a homeomorphism is that f be interior).

Regards,

Nalin Pithwa.

Some basic facts about connectedness and compactness

Reference: Hocking and Young’s Topology, Dover Publishers. Chapter 1: Topological Spaces and Functions.

Definition : Separated Space: A topological space is separated if it is the union of two disjoint, non empty, open sets.

Definition: Connected Space: A topological space is connected if it is not separated.

PS: Both separatedness and connectedness are invariant under homeomorphisms.

Lemma 1: A space is separated if and only if it is the union of two disjoint, non empty closed sets.

Lemma 2: A space S is connected if and only if the only sets in S which are both open and closed are S and the empty set.

Theorem 1: The real line E^{1} is connected.

Theorem 2: A subset X of a space S is connected if and only if there do not exist two non empty subsets A and B of X such that X = A \bigcup B, and such that (\overline{A} \bigcap B) \bigcup (A \bigcap \overline{B}) is empty.

Note the above is Prof. Rudin’s definition of connectedness.

Theorem 3: Suppose that C is a connected subset of a space S and that \{ C_{\alpha}\} os a collection of connected subsets of S, each of which intersects C. Then, S = C \bigcup (\bigcup_{\alpha}C_{\alpha}) is connected.

Corollary of above: For each n, E^{n} is connected.

Theorem 4:

Every continuous image of a connected space is connected.

Lemma 3: For n>1, the complement of the origin in E^{n} is connected.

Theorem 5: For each n>0, S^{n} is connected.

Theorem 6: If both f: S \rightarrow T and g: T \rightarrow X are continuous, then the composition gf is also continuous.

Lemma 4: A subset X of a space S is compact if and only if every covering of X by open sets in S contains a finite covering of X.

Theorem 7: A closed interval [a,b] in E^{1} is compact.

Theorem 8: Compactness is equivalent to the finite intersection property.

Theorem 9: A compact space is countably compact.

Theorem 10: Compactness and countable compactness are both invariant under continuous transformations.

Theorem 11: A closed subset of a compact space is compact.

Cheers,

Nalin Pithwa.

Some basic facts about continuity

Reference: (1) Topology by Hocking and Young especially chapter 1 (2) Analysis — Walter Rudin.

Definition 1: A transformation f: S \rightarrow T is continuous provided that if p is a limit point of a subset X of S, then f(p) is a limit point or a point of f(X).

Definition 2: We may also state the continuity requirement on f as follows: if p is a limit point of \overline{X}, then f(p) is a point of \overline{f(X)}.

Theorem 1: If S is a set with the discrete topology and f: S \rightarrow T any transformation of S into a topologized set T, then f is continuous.

Theorem 2: A real-valued function y=f(x) defined on an interval [a,b] is continuous provided that if a \leq x_{0} \leq b and \epsilon >0, then there is a number \delta >0 such that if |x-x_{0}|<\delta, x in [a,b], then |f(x) - f(x_{0})|< \epsilon. (NB: this is same as definition 1 above).

Theorem 3: Let f: S \rightarrow T be a transformation of the space S into the space T. A necessary and sufficient condition that f be continuous is that if O is any open subset of T, then its inverse image f^{-1}(O) is open in S.

Theorem 4:

A necessary and sufficient condition that the transformation f: S \rightarrow T of the space S into the space T be continuous is that if x is a point of S, and V is an open subset of T containing f(x), then there is an open set, U in S containing x and such that f(U) lies in V.

Theorem 5:

A one-to-one transformation f: S \rightarrow T of a space S onto a space T is a homeomorphism, if and only if both f and f^{-1} are continuous.

Theorem 6:

Let f: M \rightarrow N be a transformation of the metric space M with metric d into the metric space N with metric \rho. A necessary and sufficient condition that f be continuous is that if \epsilon is any positive number and x is a point of M, then there is a number \delta >0 such that if d(x,y)< \delta, then \rho(f(x), f(y)) < \epsilon.

Theorem 7:

A necessary and sufficient condition that the one-to-one mapping (that is, a continuous transformation) f: S \rightarrow T of the space S onto the space T be a homeomorphism is that f is interior. (NB: A transformation f: S \rightarrow T of the space S into the space T is said to be interior if f is continuous and if the image of every open set subset of S is open in T).

Regards,

Nalin Pithwa.

Ex: 1-7, relation between homeomorphism and continuity

Reference: Exercise: 1-7. Chapter 1. Topology by Hocking and Young.

Prove that a one-to-one transformation f: S \rightarrow T of a space S onto a space T is a homeomorphism if and only if both f and f^{-1} are continuous.

Proof:

Consider the following defintion:

Definition 1-1: A real point p is a limit point of a set X of real numbers if for any positive number \epsilon there exists a point x \in X such that 0 < |p-x| < \epsilon.

Definition of homeomorphism: A homeomorphism of S onto T is a one-to-one transformation f: S \rightarrow T which is onto and such that a point p is a limit point of a subset X of S if and only if f(p) is a limit point of f(X).

But, the above two definitions when combined mean the following: at least for the case of a Euclidean space: A real valued function y=f(x) defined on an interval [a,b] is continuous provided that if a \leq x_{0} \leq b and \epsilon >0, then there is a number \delta >0 such that if |x-x_{0}|<\delta, x in [a,b], then |f(x) - f(x_{0})|<\epsilon.

The above sub-case settles the proof for the Euclidean space E^{1}.

Now, for the more general transformation f: S \rightarrow T, consider definition 1-1 above and the following two theorems (both being equivalent definitions of continuous functions):

Theorem 1-6: Let f: S \rightarrow T be a transformation of the space S into the space T. A necessary and sufficient condition that f be continuous is that if O is any open subset of T, then its inverse image f^{-1}{O} is open in S.

Next, theorem 1-7: A necesssary and sufficient condition that the transformation f: S \rightarrow T of the space S into the space T be continuous is that if x is a point of S, and V is an open subset of T containing f(x), then there is an open set, U in S containing x such that f(U) lies in V.

Clearly, the above settles the claim for a general transformation which is one-to-onto that it is a homeomorphism if and only if both f and f^{-1} are continuous.

QED.

Your comments are welcome !

Regards,

Nalin Pithwa.

Chapter 1: Topology, Hocking and Young: A theorem due G.D. Birkhoff

Prove that the collection of all topologies on a given set S constitutes a lattice under the partial ordering of finer/coarser topology of a set.

Proof: NB. This theorem has already been proven by eminent mathematician G. D. Birkhoff.

PS: I am making my own attempt here. Your comments are most welcome !

Part 1: TST: Coarser/finer is a partial ordering of topologies.

To prove axiom 1: this order is reflexive.

Let \{ O_{\alpha}\} and \{ R_{\beta} \} be two collections of subsets of a set S, both satisfying the three defining axioms O_{1}, O_{2} and O_{3} of a topology. Let S have two topologies. We will say that the topology \mathcal{T}_{1} determined by \{ O_{\alpha} \} is a finer topology than the topology \mathcal{T}_{2} determined by \{ R_{\beta} \} if every set R_{\beta} is a union of sets O_{\alpha}, that is, each R_{\beta} is open in the \mathcal{T}_{1} topology. We will denote this situation with the symbol \mathcal{T}_{1} \geq \mathcal{T}_{2}.

Axiom 1: reflexive clearly holds because \mathcal{T}_{1} \geq \mathcal{T}_{1} as set inclusion/containment is reflexive.

Axiom 2: antisymmetric also holds true because clearly \mathcal{T}_{1} \geq \mathcal{T}_{2} and \mathcal{T}_{2} \geq \mathcal{T}_{1} together imply that \mathcal{T}_{2} = \mathcal{T}_{1}. In other words, these two topologies are equivalent, or they give rise to same basis.

Axiom 3: Transitivity holds because set inclusion/containment is transitive. If \mathcal{T}_{1} \geq \mathcal{T}_{2} and \mathcal{T}_{2} \geq \mathcal{T}_{3}, then clearly \mathcal{T}_{1} \geq \mathcal{T}_{2} \geq \mathcal{T}_{3}. In other words, there can be a chain of finer/coarser topologies for a given set.

Part 2:

TST: Under this definition of partial ordering of topologies, the partial ordering forms a lattice.

From axiom 3 proof, we know that there can exist a chain of finer/coarser topologies. So supremum and infimum can exist in a partial ordering of topologies. Hence, such a partial ordering forms a lattice.

QED.

Kindly let me know your comments and oblige about my above attempt.

Regards,

Nalin Pithwa

Solutions to Chapter 1: Topology, Hocking and Young

Exercise 1-4: Prove that the collection of all open half planes is a subbasis for the Euclidean topology of the plane.

Proof 1-4:

Note: In the Euclidean plane, we can take as a basis the collection of all interiors of squares.

Note also that a subcollection \mathcal{B} of all open sets of a topological space S is a SUBBASIS of S provided that the collection of all finite intersections of elements of $\mathcal{B}$ is a basis for S.

Clearly, from all open half planes (x > x_{i} and y > y_{i}), we can create a collection of all interior of squares.

Hence, the collection of all finite intersections of all open-half planes satisfies:

Axiom \mathcal{O_{2}}: the intersection of a finite intersection of open half planes is an open set. (interior of a square).

Axiom \mathcal{O_{1}}; (also). The union of any number of finite intersections of all open half planes is also open set (interior of a square).

Axiom O_{3}: (clearly). \phi and S are open.

QED.

Exercise 1-5:

Let S be any infinite set. Show that requiring every infinite subset of S to be open imposes the discrete topology on S.

Proof 1-5:

Case: S is countable. We neglect the subcase that a selected subcase is finite. (I am using Prof. Rudin’s definition of countable). The other subcase is that there exists a subset X_{1} \subset S, where X_{1} is countable. Let X_{1} be open. Hence, S-X_{1} is closed. But S-X_{1} is also countable. Hence, S-X_{1} is also open. Hence, there is no limit point. Hence, the topology is discrete, that is, there is no limit point.

Case: S is uncountable. Consider again a proper subset X_{1} \subset S; hence, X_{1} is open by imposition of hypothesis. Hence, S-X_{1} is closed. But, S-X_{1} \neq \phi and not finite also. Hence, S-X_{1} is infinite. Hence, S-X_{1} is open. Hence, there are no limit points. Hence, it is a discrete topology in this case also.

QED.

Your comments/observations are welcome !

Regards,

Nalin Pithwa.