Topology 2nd note

1. Basis

Eg

\({\mathbb{R}},\mathcal{B}:\left\{ (a,b),(a,b) - K \right\},K = \left\{ \frac{1}{n}~|~n \in {\mathbb{Z}}_{+} \right\}\)

\(\mathcal{B}\) is a basis.

\((a,b) \cap \left( (c,d) - K \right) = \left( \max\left\{ a,c \right\},\min\left\{ b,d \right\} \right) - K\)

\(\left( (a,b) - K \right) \cap \left( (c,d) - K \right) = \left( \max\left\{ a,c \right\},\min\left\{ b,d \right\} \right) - K\)

\(\mathcal{B}\) generate the \(K\)-topology in \(\mathbb{R}\).

Eg

\({\mathbb{R}},\mathcal{B} = \{\lbrack n,n + 1)~|~n \in {\mathbb{Z}}\}\) is a basis.

Eg

\(X = {\mathbb{R}}^{n},x = \left( x_{1},\ldots,x_{n} \right),y = \left( y_{1},\ldots,y_{n} \right)\)

\(d(x,y) = \sqrt{\left( x_{1} - y_{1} \right)^{2} + \ldots{+ \left( x_{n} - y_{n} \right)}^{2}},B(x,\varepsilon) = \left\{ y \in {\mathbb{R}}^{n}~|~d(x,y) < \varepsilon \right\}\)

\(\mathcal{B}_{1} = \left\{ B(x,\varepsilon) \subset {\mathbb{R}}^{n}~|~x \in {\mathbb{R}}^{n},\varepsilon > 0 \right\}\)

Lemma

\(\mathcal{B}_{1}\) is a basis on \({\mathbb{R}}^{n}\).

这里少张图

\(\mathcal{B}_{1}\) generates the standard topology on \({\mathbb{R}}^{n}\) denoted by \({\mathbb{E}}^{n}\). Euclidean Space.

\(\mathcal{B}_{2} = \left\{ \left( a_{1},b_{1} \right) \times \ldots \times \left( a_{n},b_{n} \right)~|~\left( a_{i},b_{i} \right) \subset {\mathbb{R}} \right\}\)

Lemma

\(\mathcal{B}_{2}\) is a basis on \({\mathbb{R}}_{n}\).

这里少张图

\(\begin{aligned} & \left( \left( a_{1},b_{1} \right) \times \ldots \times \left( a_{n},b_{n} \right) \right) \cap \left( \left( c_{1},d_{1} \right) \times \ldots \times \left( c_{n},d_{n} \right) \right) \\ = & \left( \left( a_{1},b_{1} \right) \cap \left( c_{1},d_{1} \right) \times \ldots \times \left( a_{n},b_{n} \right) \cap \left( c_{n},d_{n} \right) \right) \in \mathcal{B}_{2} \end{aligned}\)

Question

Does \(\mathcal{B}_{2}\) generate the standard topology on \({\mathbb{R}}^{n}\) ?

Proposition

\(\left( X,\mathcal{T} \right)\) is a topol. space, and \(\mathcal{C}\) is a collectopn of open sets in \(X\). If \(\forall\) open set \(U \subset X\) and \(\forall x \in U,\exists\) an open set \(V \in \mathcal{C}\), s.t. \(\underline{x \in V \subset U}\). Then \(\mathcal{C}\) is a basis that generates the given topology \(\mathcal{T}\).

Topology \(\rightsquigarrow\) Basis

Pf

1. We check that \(\mathcal{C}\) is a basis.

   1. \(\forall x \in X,X \in \mathcal{T}\), by assumption, \(\exists\) an open set \(V\) in \(\mathcal{C}\), s.t. \(x \in V \subset X\).

   2. Suppose \(V_{1},V_{2} \in \mathcal{C}\). \(V_{1} \cap V_{2}\) is open. \(\forall x \in V_{1} \cap V_{2}\), by assumption, \(\exists V_{3} \in \mathcal{C}\), s.t. \(x \in V_{3} \subset V_{1} \cap V_{2}\).

   So \(\mathcal{C}\) is a basis.

2. Suppose \(\mathcal{C}\) generates the topology \(T_{1}\). We check that \(\mathcal{T}_{1} = \mathcal{T}\).

   • "\(\subset\)"

     \(\forall W \in \mathcal{T_{1}},W\) is a union of subsets in \(\mathcal{C}\), all of which is open in \(\mathcal{T}\). So \(W \in \mathcal{T}\). Thus \(\mathcal{T_{1}} \subset \mathcal{T}\).

   • "\(\supset\)"

     \(\forall U \in \mathcal{T}\). \(\forall x \in U\), by assumption, \(\exists V \in \mathcal{C}\), s.t. \(x \in V \subset U\). By previous lemma, \(U \in \mathcal{T_{1}}\). Thus \(\mathcal{T} \subset \mathcal{T}_{1}\).

   therefore, \(\mathcal{T}_{1} = \mathcal{T}\).  \(\blacksquare\)

Answer to the Question: Yes!

1. \(\left( a_{1},b_{1} \right) \times \ldots \times \left( a_{n},b_{n} \right)\) is open in \({\mathbb{E}}^{n}\).

   这里少张图

2. 这里少张图

   \(U\) open in \({\mathbb{E}}^{n},\forall x \in U,\exists V \in \mathcal{B}_{2}\), s.t. \(x \in V \subset U\).

2. Subspace

\(\left( X,\mathcal{T} \right)\) is a topol. space. \(A \subset X\). \(\mathcal{T}_{A} = \left\{ U \cap A~|~U \in \mathcal{T} \right\}\)

Proposition

\(\mathcal{T}_{A}\) is a topology on \(A\).

Pf

1. \(\varnothing \in \mathcal{T}_{A}(\varnothing = \varnothing \cap A),A \in \mathcal{T}_{A}(A = X \cap A).\)

2. \(U_{\lambda} \cap A,\lambda \in \Lambda\). \(U_{\lambda} \in \mathcal{T}.\) \(\bigcup_{(\lambda \in \Lambda)\left( U_{\lambda} \cap A \right)} = \left( \bigcup_{\lambda \in \Lambda}U_{\lambda} \right) \cap A \in \mathcal{T}_{A}\). Since \(\bigcup_{\lambda \in \Lambda}U_{\lambda} \in \mathcal{T}\).

3. \(\left( U_{1} \cap A \right) \cap \left( U_{2} \cap A \right) = \left( U_{1} \cap U_{2} \right) \cap A \in \mathcal{T}_{A},\) since \(U_{1} \cap U_{2} \in T\).  \(\blacksquare\)

Def

\(\left( A,\mathcal{T}_{A} \right)\) is called a subspace of \(\left( X,\mathcal{T} \right)\).

\(\mathcal{T}_{A}\) is called the subspace (or induced) topology on \(A\).

Eg

\({\mathbb{Z}} \subset {\mathbb{R}}_{\text{std}},\forall z \in {\mathbb{Z}},\left\{ z \right\} = \left( z - \frac{1}{2},z + \frac{1}{2} \right) \cap {\mathbb{Z}}\) is open.

The induced topology on \(\mathbb{Z}\) is discrete.

Eg

\(\lbrack 0,1\rbrack \subset {\mathbb{R}}_{\text{std}}\)

\(\lbrack 0,0.3),(0.7,1\rbrack\) are open in the subspace topology.

Lemma

\(\left( X,\mathcal{T} \right)\) is a topol. space, \(\mathcal{B}\) is a basis generate \(\mathcal{T}\).

\(A \subset X,\mathcal{B}_{A} = \left\{ B \cap A~|~B \in \mathcal{B} \right\}\). Then \(\mathcal{B}_{A}\) is a basis on \(A\) generating the subspace topology \(\mathcal{T}_{A}\).

Pf

1. \(\forall B \in \mathcal{B},B\) is open in \(\mathcal{T}\). So \(B \cap A \in \mathcal{T}_{A}\).

2. \(\forall U \cap A \in \mathcal{T}_{A}\), where \(U \in \mathcal{T}\). \(\forall x \in U \cap A,\exists B \in \mathcal{B}\), s.t. \(x \in B \subset U.\) Then \(x \in B \cap A \subset U \cap A\), where \(B \cap A \in \mathcal{B}_{A}\).

By previous proposition, the lemma holds.   \(\blacksquare\)

Eg

\(S^{1} =\) 图片 \(\subset {\mathbb{E}}^{2}\).

Eg

\(S^{2} =\) 图片 \(\subset {\mathbb{E}}^{3}\).

Lemma

\(\left( X,\mathcal{T} \right)\) is a topol. space, \(B \subset A \subset X\). Let \(\left( A,\mathcal{T}_{A} \right)\) and \(\left( B,\mathcal{T}_{B} \right)\) be the subspaces of \(\left( X,\mathcal{T} \right)\), and \(\left( B{,\left( \mathcal{T}_{A} \right)}_{B} \right)\) be the subspace of \(\left( A,\mathcal{T}_{A} \right)\). Then \(\left( \mathcal{T}_{A} \right)_{B} = \mathcal{T}_{B}\).

Pf

• "\(\subset\)"

  \(\forall U \in \left( \mathcal{T}_{A} \right)_{B}\), then \(\exists\) an open set \(U_{1} \in \mathcal{T}_{A},\) s.t. \(U = U_{1} \cap B\).

  Since \(U_{1} \in \mathcal{T}_{A},\exists\) an open set \(U_{2} \in \mathcal{T},\) s.t. \(U_{1} = U_{2} \cap A\). So \(U = \left( U_{2} \cap A \right) \cap B = U_{2} \cap B\). Therefore \(U \in \mathcal{T}_{B}\).

• "\(\supset\)"

  \(\forall U \in \mathcal{T}_{B},U = U_{2} \cap B\), where \(U_{2} \in \mathcal{T}\). \(U = \left( U_{2} \cap A \right) \cap B\), where \(U_{2} \cap A \in \mathcal{T}_{A}\). So \(U \in \left( \mathcal{T}_{A} \right)_{B}\).  \(\blacksquare\)

3. Product Space

Def

\(X,Y:\) topol. spaces. \(X \times Y:\) Cartesian product.

\(\mathcal{B} = \left\{ U \times V~|~U\text{ is open in }X,V\text{ is open in }Y \right\}\)

Lemma

\(\mathcal{B}\) is a basis on \(X \times Y\).

Pf

1. \(\forall(x,y) \in X \times Y,(x,y) \in X \times Y \in \mathcal{B}\).

2. \(U_{1} \times V_{1},U_{2} \times V_{2} \in \mathcal{B}.\left( U_{1} \times V_{1} \right) \cap \left( U_{2} \times V_{2} \right) = \left( U_{1} \cap U_{2} \right) \times \left( V_{1} \cap V_{2} \right) \in \mathcal{B}\)

   这里少张图

   So \(\mathcal{B}\) is a basis on \(X \times Y\).  \(\blacksquare\)

The topology generated by \(\mathcal{B}\) is called the product topology on \(X \times Y\).

Lemma

If \(\mathcal{B}_{1}\) is a basis for \(X\), and \(\mathcal{B}_{2}\) is a basis for \(Y\), then \(\mathcal{B}_{3} = \left\{ B_{1} \times B_{2}~|~B_{i} \in \mathcal{B}_{i} \right\}\) is a basis generating the product topology on \(X \times Y\).

Pf

1. \(\forall B_{1} \times B_{2} \in \mathcal{B}_{3},B_{1} \times B_{2}\) is open in \(X \times Y\).

2. Suppose \(W\) is open in \(X \times Y.\forall(x,y) \in W.\) By definition, \(\exists U\left( \text{open in }X \right),V\left( \text{open in }Y \right),\) s.t. \((x,y) \in U \times V \subset W\).

   这里少张图

   \(\exists B_{1} \in \mathcal{B}_{1},\) s.t. \(x \in B_{1} \subset U\)

   \(\exists B_{2} \in \mathcal{B}_{2},\) s.t. \(x \in B_{2} \subset V\)

   Hence \((x,y) \in B_{1} \times B_{2} \subset U \times V \subset W\).

By a previous proposition, the lemma holds.  \(\blacksquare\)

Eg

\(\begin{aligned} & {\mathbb{E}}^{2} & = & {\mathbb{E}}^{1} & \times & {\mathbb{E}}^{1} \\ & \cup & & \cup & & \cup \\ & (a,b) \times (c,d) & & (a,b) & & (c,d) \end{aligned}\): basis elemts.

Lemma

\(X,Y:\) topol.spaces. \(A \subset X,B \subset Y.\) Then the induced topology on \(A \times B\) as a subspace of \(X \times Y\) is the same as the product topology on \(A \times B\), where \(A,B\) are subspaces of \(X,Y\), respectively.

Pf

\(\mathcal{B} = \left\{ (U \cap A) \times (V \cap B)~|~U\text{ is open in }X,V\text{ is open in }Y \right\}\) generates teh product topol.

We check that \(\mathcal{B}\) generates the induced topol. on \(A \times B\) as well.

1. \((U \cap A) \times (V \cap B) = (U \times V) \cap (A \times B)\).

   Since \(U \times V\) is open in \(X \times Y,(U \cap A) \times (V \cap B)\) is open in the the induced topology on \(A \times B\).

2. Suppose \(W\) is open in the induced topol. on \(A \times B\).

   Then \(\exists W_{1}\)(open in \(X \times Y\)), s.t. \(W = W_{1} \cap (A \times B)\).

   \(\forall(x,y) \in W = W_{1} \cap (A \times B) \subset W_{1},\exists U_{1}\left( \text{open in }X \right),V_{1}\left( \text{open in }Y \right),\) s.t. \((x,y) \in U_{1} \times V_{1} \subset W_{1}\).

   So \((x,y) \in \underset{= \left( U_{1} \cap A \right) \times \left( V_{1} \cap B \right) \in B}{\left( U_{1} \times V_{1} \right) \cap (A \times B)} \subset W_{1} \cap (A \times B) = W\)   \(\blacksquare\)

Eg

\(S^{1} \subset {\mathbb{E}}^{2},\lbrack 0,1\rbrack \subset {\mathbb{E}}^{1},S^{1} \times \lbrack 0,1\rbrack \subset {\mathbb{E}}^{2} \times {\mathbb{E}}^{1} = {\mathbb{E}}^{3}\)

这里少张图

Cylinder

Eg

\(S^{1} \subset {\mathbb{E}}^{2},S^{1} \subset {\mathbb{E}}^{2},S^{1} \times S^{1} \subset {\mathbb{E}}^{2} \times {\mathbb{E}}^{2} = {\mathbb{E}}^{4}\)

这里少张图

Torus

Def

product topology on \(X_{1} \times X_{2} \times \ldots \times X_{n}\), where \(X_{i}\) is a topol. space.

\(\mathcal{B} = \left\{ U_{1} \times U_{2} \times \ldots \times U_{n}~|~U_{i}\text{ is open in }X_{i} \right\}\) is a basis.

Def

\(\begin{aligned} p_{i}:X_{1} \times X_{2} \times \ldots \times X_{n} & \rightarrow X_{i} \\ x_{1} \times x_{2} \times \ldots \times x_{n} & \mapsto x_{i} \end{aligned}\)

is a projective.

4. Closed set

Def

\(X\) is a topol. space, \(C \subset X\). \(C\) is called a closed subset if \(X - C\) is open.

Eg

\({\mathbb{E}}^{1},\lbrack a,b\rbrack\) is closed; \(( - \infty,b\rbrack\) is closed.

\({\mathbb{E}}^{2},\overline{B(x,\varepsilon)} ≔ \left\{ y \in {\mathbb{E}}^{2}~|~d(x,y) \leq \varepsilon \right\}\) is closed.

Eg

\({\mathbb{R}}_{\text{fc}}\) finite complement topology

\(C \subset {\mathbb{R}}_{\text{fc}}\) is closed \(\Leftrightarrow C = {\mathbb{R}}\) or finite subsets.

Proposition

\(X\) is a topol. space, then

1. \(\varnothing\) and \(X\) are closed;

2. Arbitrary intersection of closed sets is closed;

   \(C_{\lambda},\lambda \in \Lambda\) is closed in \(X\). \(X - \bigcap_{\lambda \in \Lambda}C_{\lambda} = \bigcup_{(\lambda \in \Lambda)\left( X - C_{\lambda} \right)}\) is open in \(X\). So \(\bigcap_{\lambda \in \Lambda}C_{\lambda}\) is closed in \(X\).

3. The union of finitely many closed sets is closed.

   \(C_{1},C_{2}\) is closed in \(X\), \(X - \left( C_{1} \cup C_{2} \right) = \left( X - C_{1} \right) \cap \left( X - C_{2} \right)\) is open in \(X\). So \(C_{1} \cup C_{2}\) is closed in \(X\).

Eg

这里缺少一个讲义存在但是课上没讲的例子

Eg

\({\mathbb{E}}^{n}\), a single point set \(\left\{ p_{t} \right\}\) is closed.

\({\mathbb{E}}^{n} - \left\{ p_{t} \right\}\) is open. \(x \in {\mathbb{E}}^{n} - \left\{ p_{t} \right\},\exists\varepsilon > 0\), s.t. \(x \in B(x,\varepsilon) \subset {\mathbb{E}}^{n} - \left\{ p_{t} \right\}\).

Eg

\({\mathbb{R}},\mathcal{B} = \{\lbrack n,n + 1)~|~n \in {\mathbb{Z}}\}\) generates a topology \(\mathcal{T}\).

Then \(\left\{ 0 \right\}\) is not closed in \(\mathcal{T}\).

Pf

Suppose \({\mathbb{R}}\backslash\left\{ 0 \right\} = ( - \infty,0) \cup (0, + \infty)\) is open. \(\frac{1}{2} \in {\mathbb{R}}\backslash\left\{ 0 \right\}\), then \(\exists\) a basis element \(\lbrack n,n + 1),n \in {\mathbb{Z}}\), s.t. \(\frac{1}{2} \in \lbrack n,n + 1) \subset {\mathbb{R}}\backslash\left\{ 0 \right\}\).

No such \(n\) exists. Contradction!

Lemma

\(\left( X,\mathcal{T} \right):\) topol. space, \(A \subset X,\left( A,\mathcal{T}_{A} \right):\) subspace.

Then \(C \subset A\) is closed in \(\left( A,\mathcal{T}_{A} \right) \Leftrightarrow \exists\) a closed set \(C_{1}\) in \(X\), s.t. \(C = C_{1} \cap A\).

Pf

\(\begin{aligned} & C\text{ is closed in }\left( A,\mathcal{T}_{A} \right) \\ \Leftrightarrow & A - C\text{ is open in }\left( A,\mathcal{T}_{A} \right) \\ \Leftrightarrow & \exists\text{ an open set }U_{1}\text{ in }X,\text{ s.t. }A - C = U_{1} \cap A\text{ i.e. }C = \left( X - U_{1} \right) \cap A \end{aligned}\)

这里少张图

Here \(X - U_{1}\) is closed in \(X\). Set \(C_{1} = X - U_{1}\). Then \(C = C_{1} \cap A\).  \(\blacksquare\)

5. Hausdorff

Def

A topol. space \(X\) is Hausdorff if \(\forall\) any pair of distant points \(x\) and \(y\), \(\exists\) disjoint nbhds \(U\) and \(V\) of \(x\) and \(y\) respectively.

这里少张图

\(U \cap V = \varnothing\)

Eg

\({\mathbb{E}}^{n}\) is Hausdorff.

Eg

\(X\) with trivial topology is not Hausdorff, where \(X\) has at least \(2\) elements.

Lemma

If \(X\) is Hausdorff, then every single point set is closed.

Pf

\(\forall\left\{ x \right\} \subset X\), we show that \(X\backslash\left\{ x \right\}\) is open. \(\forall y \in X\backslash\left\{ x \right\},y \neq x\). Since \(X\) is Hausdorff, \(\exists\) nbhd \(U_{y}\) of \(y\), nbhd \(U_{x}\) of \(x\), s.t. \(U_{x} \cap U_{y} = \varnothing\). \(y \in U_{y} \subset x\backslash\left\{ x \right\}\). So \(X\backslash\left\{ x \right\}\) is open, and \(\left\{ x \right\}\) is closed.

Eg

\({\mathbb{R}},\mathcal{B} = \{\lbrack n,n + 1)~|~n \in {\mathbb{Z}}\}\) generates \(\mathcal{T},\mathcal{T}\) is not Hausdorff.

Eg

\({\mathbb{R}}_{\text{fc}}\) is not Hausdorff.(every single point set is closed)

\(x \neq y \in {\mathbb{R}}\). Suppose \(\exists\) nbhds \(U_{x}\) of \(x\), \(U_{y}\) of \(y\), s.t. \(U_{x} \cap U_{y} = \varnothing\), them \(\underset{\text{infinite}}{U_{y}} \subset \underset{\text{finite}}{{\mathbb{R}} - U_{x}}\). Contradction!   \(\blacksquare\)

HW

\(\begin{array}{l} §13:5,6,7,8 \\ §16:3,4,6 \\ §17:2,3,4 \end{array}\)