# 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.

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.

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.