Topological Space¶

Definition of Topology

Suppose set \(X\), and a family set \(\tau\) of \(X\) is said to be a topology if

(i) \(\varnothing, X\subset \tau\),

(ii) if \(\{U_\lambda\}_{\lambda\in\Lambda}\subset \tau\), then \(\bigcup\limits_{\lambda\in \Lambda} U_\lambda\in \tau\),

(iii) if \(\{U_n\}_{1\leq n\leq N} \subset \tau\), then \(\bigcap\limits_{1\leq n\leq N}U_n\in \tau\).

The element of \(\tau\) is called an open set.

Example. Two common topologies on a set \(X\).

(i) discrete topology \(2^{X}\).

(ii) indiscrete/trivial topology \(\{\varnothing, X\}\).

Example. The topologies on \(\mathbb{R}\).

(i) Finite-complement topology

\[ \tau_f=\{A\subset \mathbb{R}: |A^c|<\infty\} \cup \{\varnothing\} \]

(ii) Countable-complement topology

\[ \tau_c = \{A\subset \mathbb{R}: |A^c| \text{ is countable}\} \cup \{\varnothing\} \]

(iii) Euclidean topology

\[ \tau_e = \{A\subset \mathbb{R}: A=\bigcup_{\lambda\in\Lambda} I_\lambda, \Lambda \subset \mathbb{R}\} \]

where \(\Lambda\) is an index set and compose maybe finite/infinite/zero number of elements. \(I_\lambda\) is an arbitrary open interval. We denote the space as \(\mathbb{E}^1=(\mathbb{R}, \tau_e)\).

Proof

(i) and (ii) holds for condition two and three because De Morgan's formula

\[ \left(\bigcup_{\lambda\in\Lambda}A_\lambda\right)^c=\bigcap_{\lambda\in\Lambda}A_\lambda^c,\quad \left(\bigcap_{n=1}^N A_n\right)^c=\bigcup_{n=1}^N A_n^c. \]

Comparison of topologies

Suppose \(\tau\), \(\tau'\) are two topologies on \(X\). We say \(\tau'\) is finer than \(\tau\) (or bigger), if

\[ \tau\subset \tau'. \]

And at this time the two topologies is comparable.

Basis for a topology¶

Basis for a topology

A family set \(\mathcal{B}\) of \(X\) is said to be a topological basis, if

(i) \(\forall x\in X\), \(\exists B\in\mathcal{B}\) s.t. \(x\in B\). Or equivalently, \(X=\bigcup\limits_{B\in \mathcal{B}}B\).

(ii) \(\forall B_1,B_2\in \mathcal{B}\), \(\forall x\in B_1\cap B_2\), \(\exists B\in\mathcal{B}\) such that \(x\in B\) and \(B\subset B_1\cap B_2\).

We now define a collection of subsets of \(X\) by a basis for a topology, which turns out to be a topology.

Topology Generated by basis

A collection \(\tau_B\) of subsets of \(X\) is defined as follows. A subset \(U\subset X\) belongs to \(\tau_B\), if \(\forall x\in U\), \(\exists B\in \mathcal{B}\) s.t. \(x\in B\) and \(B\subset U\). Then \(\tau_B\) is a topology on \(X\).

Proof

Check this by definition. Note that \(\varnothing\in \tau\) and \(X\in \tau\) (by def (i) of basis for a topology).

(ii) For \(\{U_\lambda\}_{\lambda\in\Lambda}\subset \tau\), by definition, \(\exists B_\lambda\in\mathcal{B}\), s.t. \(\forall x\in U_\lambda\), \(x\in B_\lambda\) and \(B_\lambda\in U_\lambda\). So for

\[ U:=\bigcup_{\lambda\in\Lambda} U_\lambda \]

\(x\in U\), \(\exists \lambda_0\) such that \(x\in U_{\lambda_0}\), then \(\exists B_{\lambda_0}\subset U_{\lambda_0}\subset U\) such that \(x\in B_\lambda\).

(iii) Use the condition (ii) of the definition. We only prove if \(U_1,U_2\in \tau\), then \(U_1\cap U_2 \in \tau\).

Generation by a union of basis for a topology

Suppose \(\tau\) is a topology space generated by \(\mathcal{B}\), then \(\forall U\in \tau\), there exsits \(\mathcal{B}_1\subset \mathcal{B}\) such that

\[ U=\bigcup_{B\in\mathcal{B}_1} B. \]

Conversely, a collection \(\tau\) defined by the above expression is a topology generated by basis \(\mathcal{B}\).

Proof

Sufficient. Easy to check by definition.
necessary.

For each \(U\in \tau\), by definition, for each \(x\in U\), there exists \(B_x\in \mathcal{B}\), s.t. \(x\in B_x\subset U\).

So choose \(\mathcal{B}_1=\{B_x: x\in U\}\), and we have

\[ U = \bigcup_{B\in\mathcal{B}_1}B=\bigcup_{x\in U}B_x. \]

\(\square\)

This expression for \(U\) is not unique.

Given a topology \(\tau\), we could check whether it is generated by a specific topological basis.

Lemma: find a topological basis from a topology

Suppose \(\tau\) is a topology on \(X\), a collection \(\mathcal{C}\) of (open) subsets of \(X\) is said to be a topological basis of \(\tau\), if for each open set \(U\) and each \(x\in U\), there exsits \(C\in\mathcal{C}\) such that \(x\in C\subset U\).

Actually, this topology is the same as the topology generated by basis \(\mathcal{C}\).

Proof

We shall check \(\mathcal{C}\) is a basis for a topology.
Let \(\tau'\) to denote the topology generated by the basis \(\mathcal{C}\). By definition, we have \(\tau'\subset \tau\) (Readers could check). For the other direction, we have for each open set \(U\), choose \(\mathcal{C}_1=\{C_x: x\in C_x\subset \mathcal{C}\}\), then

\[ U=\bigcup_{C \in \mathcal{C}_1}C\subset \tau'. \]

Now we could give a criterion for comparing two topologies generated by two basis.

Lemma: criterion for comparing two topologies generated by two basis

Suppose \(\mathcal{B}\) and \(\mathcal{B}'\) are two basis for topologies \(\tau\) and \(\tau'\), respectively. Then \(\tau'\) is finer than \(\tau\), iff for each \(x\in X\) and \(B\in\mathcal{B}\) with \(x\in B\), there exists \(B'\in\mathcal{B}'\) such that \(x\in B'\subset B\).

Proof

Show by definition.

\(\square\)

Apply the above lemma to show the following result.

Example. Ball basis and triangular basis generate the same topology on \(\mathbb{R}^2\).

Product Topology¶

Definition of product topology on \(X\times Y\)

Suppose \(X\) and \(Y\) are topological spaces, the product topology on \(X\times Y\) is a collection of subsets of \(X\) generated by basis \(\mathcal{B}\), where \(\mathcal{B}=\{U\times V: U\in\tau_X, V\in \tau_Y\}\).

We shall show that the above \(\mathcal{B}\) is a topological basis.

Proof

using

\[ (U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2). \]

Note that \(\mathcal{B}\) itself is not a topology of \(X\times Y\).

When does a collection of subsets of \(X\times Y\) becomes a basis for the topology on \(X\times Y\)?

Basis for a topology on \(X\times Y\)

Suppose \(\mathcal{B}\), \(\mathcal{C}\) are topological basis on \(X\) and \(Y\), respectively. Then

\[ \mathcal{D}=\{B\times C: B\in \mathcal{B}, C\in\mathcal{C}\} \]

is a basis for a topology on \(X\times Y\).

Proof

Now we make use of projection function to show some relationships.

Projections

Suppose \(X\) and \(Y\) are two non-empty sets. Define a map \(\pi_1:X\times Y\rightarrow X\) by

\[ \pi_1(x,y)=x \]

and \(\pi_2: X\times Y\rightarrow Y\) by

\[ \pi_2(x,y)=y. \]

They are called the projections of \(X\times Y\) onto its first and second factor, respectively.

Projections are surjective by definition. We would show that it is continuous naturally in the following chapter.

topology generated by projections

Collection

\[ \mathcal{S}=\{\pi_1^{-1} (U): U \text{ is open in } X\}\cup \{\pi_2^{-1} (V): V \text{ is open in } Y\} \]

is a sub-basis of the product topology of \(X\times Y\).

Proof

By

\[ (U\times V)=(U\times Y) \cap (X\times V). \]

Show that product topology is equal to the topology generated by \(\mathcal{S}\).

Subspace Topology¶

Definition of Subspace topology

Suppose \(X\) is a topological space with \(\tau\). If \(Y\subset X\), then the collection

\[ \tau_Y=\{U\cap Y: U\in\tau\}. \]

is a topology on \(Y\), called the subspace topology. With this topology, \(Y\) is called a subspace of \(X\).

Proof

Easy to have

\[ \varnothing=\varnothing\cap Y,\quad Y=X\cap Y. \]

and for condition two

\[ \bigcup_{\lambda\in\Lambda}(U_\lambda \cap Y) = \left(\bigcup_{\lambda\in\Lambda}U_\lambda \right)\cap Y \]

\[ \bigcap_{n=1}^N(U_n \cap Y) = \left(\bigcap_{n=1}^N U_n \right)\cap Y \]

\(\square\)

Quotient Topology¶

Definition of quotient map

Suppose \(X\) and \(Y\) are topological spaces, \(p:X\rightarrow Y\) is called a quotient map if

(i) \(p\) is surjective,

(ii) \(V\subset Y\) is an open set in \(Y\), iff \(f^{-1}(V)\) is open in \(X\).

The (ii) could be partitioned into two conditions: \(p\) is continuous, and for \(V\subset Y\), \(p^{-1}(V)\) is open in \(X\), implies \(V\) is open in \(Y\).

Now we use quotient map to construct a topology, namely quotient topology.

Definition of quotient topology

Suppose \(X\) is a topological space, \(A\) is a set and \(p:X\rightarrow A\) is a surjective map. Then there exsits exactly one topology \(\tau\) on \(A\) such that \(p\) becomes a quotient map. It is called the quotient topology induced by \(p\).

Proof

The open set \(U\in\tau\) such that \(p^{-1}(U)\) is open in \(X\). This guarantees the continuity of \(p\), and actually makes the topology the finest topology that makes \(p\) continuous. We claim that this construction gives a topology.

We shall show that \(\tau\) is a topology. Apparently, \(\varnothing\) and \(A\) are in \(\tau\) since \(p^{-1}(\varnothing)=\varnothing\), \(p^{-1}(A)=X\). And

\[ p^{-1}\left(\bigcup_{\alpha\in A}U_\alpha\right)=\bigcup_{\alpha\in A} p^{-1}(U_\alpha),\quad p^{-1}\left(\bigcap_{n=1}^N U_n\right)=\bigcap_{n=1}^N p^{-1}(U_n) \]

A special case for quotient topology is defined as follows.

Special case of quotient space

Suppose \(X\) is a topological space, \(X^*\) is a partition of \(X\) into disjoint subsets whose union is \(X\). Let \(p:X\rightarrow X^*\) be a surjective map that carries each point of \(X\) into the element of \(X^*\) containing it. In the quotient topology induced by \(p\), \(X^*\) is called quotient space of \(X\).

We can describe the topology of \(X^∗\) in another way. A subset \(U\) of \(X^∗\) is a collection of equivalence classes, and the set \(p^{−1}(U)\) is just the union of the equivalence classes belonging to \(U\). Thus the typical open set of \(X^∗\) is a collection of equivalence classes whose union is an open set of \(X\).

Continuous functions¶

Definition of Continuous functions

Suppose \(X\), \(Y\) are topological spaces, \(f:X\rightarrow Y\) is a map. \(f\) is said to be continuous, if for all \(U\in \tau_Y\), we have \(f^{-1}(U)\in \tau_X\).

Note that for \(V\subset Y\), if \(V\cap f(X)=\varnothing\), then \(f^{-1}(V)=\varnothing\).

Example. Suppose \(f:X\rightarrow Y\) is a bijection. Show that the following statements are equivalent.

(i) \(f^{-1}\) is continuous,

(ii) \(f\) is open,

(iii) \(f\) is closed.

Proof

(i) implies (ii) is by definition.

(ii) implies (iii) is by \(f(U^c)=[f(U)]^c\).

Rules for constructing continuous functions

Suppose \(X, Y, Z\) are topological spaces.

(i)

Proof

Pasting lemma (黏结引理)

Maps into products

Assume \(f: A\rightarrow X\times Y\) is defined by

\[ f(a)=(f_1(a), f_2(a)). \]

Then \(f\) is continuous iff \(f_1\) and \(f_2\) are continuous.

Here \(f_1, f_2\) are called the coordinate functions of \(f\).

Note that if the domain is product space, then the aboe proposition is not ture.

Example. Suppose \(F: \mathbb{E}\times \mathbb{E}\rightarrow \mathbb{E}\) defined by

\[ F(x,y)=\begin{cases} xy/(x^2+y^2),\quad &(x,y)\neq (0,0),\\ 0, \quad & (x,y)=(0,0). \end{cases} \]

Show that \(F_1, F_2\) is continuous but \(F\) itself is not continuous.

Homeomorphism¶

Definition of Hemeomorphism

Denoted by \(\simeq\). Suppose \(X\), \(Y\) is a topological space, and \(f:X\rightarrow Y\) is a map. \(f\) is called homeomorphism, if \(f\) is bijective, \(f\) and \(f^{-1}\) are both continuous.

For given \(X\) and \(Y\), we call them homeomorphic if there exsits a homeomorphism \(f:X\rightarrow Y\).

Example. Some homeomorphisms.

(i) open intervals is homeomorphic with \(\mathbb{E}^1\).

(ii) open unit balls in \(\mathbb{E}^n\) is homeomorphic with \(\mathbb{E}^n\).

(iii) Any convex polygon is homeomorphic with a unit disc.

(iv) \(z\mapsto z+1\) is analytic at \(\infty\).

Proof

(i) using \(x\mapsto \tan x\) from \((-\pi/2, \pi/2)\) to \(\mathbb{R}\).

(ii) using \(x\mapsto x+\frac{x}{\|x\|}\).

Topological imbedding

Suppose \(f:X\rightarrow Y\) is a continuous and injective map. If restricted on \(f(X)\), \(f\) is a hemeomorphism, then we call \(f\) is an embedding of \(X\) in \(Y\).

Example.

(i) \(\mathbb{R}^m\rightarrow \mathbb{R}^n\), \(m < n\), defined by

\[ (x_1,\cdots,x_m)\mapsto (x_1,\cdots,x_m,0,\cdots, 0) \]

Example. Suppose \(X\simeq X'\), \(Y\simeq Y'\), then \(X\times Y\simeq X'\times Y'\).

Proof

From a homeomorphism \(f:X\rightarrow X'\), \(g:Y\rightarrow Y'\), define

\[ (f\times g)(x,y)=(f(x),g(y)). \]

which is a bijection, whose inverse is \(f^{-1}\times g^{-1}\).

Prove \(f\times g\) is continuous.

Example. Heegaard decomposition of \(S^3\).

(i) \(S^3 - S^1\simeq \mathbb{R}^3-\mathbb{R}\),

Example. Assume \(f: X\rightarrow Y\) is a bijection, then the following statement is equivalent.

(i) \(f\) is open.

(ii) \(f\) is closed.

(iii) \(f^{-1}\) is continuous.