\(X\) metric space

Recall \(X\) is sequentially compact if \(∀(x_n)∃x_{n_k}:(x_{n_k})\) converges.

**Theorem. **\(X\) is compact\(⇒X\) is sequentially compact.

**Proof. **Seen in Metric Spaces. ◻

**Theorem. **\(X\) is sequentially compact\(⇒X\) is compact

Proof relies on 2 lemmas:

**Lemma 1.** **(Lebesgue)**Let \(X\) be a sequentially compact metric space. For any open cover 𝒰 of \(X\) there is some \(ε>0\) such that \(∀x∈X∃U∈𝒰: B (x,ε)⊂U\).

**Proof. **Done last time. ◻

**Definition. **The ε of this lemma is called a **Lebesgue number** of the cover 𝒰.

**Definition. **Given \(ε>0\), an **ε-net** of a metric space is a set \(N⊂X\) such that \(\{B (n,ε):n∈N\}\) is a cover of \(X\).

**Example.**

- \(ℤ^2⊂ℝ^2\) is a \(\sqrt{2}\)-cover of \(ℝ^2\)
- \(ℤ⊂ℝ\) is a 1-cover of \(ℝ\)

**Lemma 2. **Let \(X\) be a sequentially compact metric space, and let \(ε>0\). Then \(\exists\) finite ε-net for \(X\).

**Proof. **Seen in Metric Spaces. ◻

**Proof of Theorem.** Let 𝒰 be a cover of \(X\). Since \(X\) is sequentially compact by Lemma 1, 𝒰 has a Lebesgue number \(ε>0\) (ie. \(∀x, B (x,ε)⊂U\) for some \(U⊂𝒰\))

By Lemma 2 \(\exists\) finite ε-net \(N=\{x_1, …,x_n\}\). So \(X=B (x_1,ε)∪⋯∪B (x_n,ε), B (x_i,ε)⊂U_i∈𝒰\).

So \(X=\bigcup_{i=1}^n U_i\) i.e. 𝒰 has a finite subcover\(⇒X\) compact.

We use quotient spaces to

- visualize spaces that are hard to see
- define new spaces

Mathematician's glue

**Definition. **For a set \(X\), a relation ℛ is a subset of \(X×X\).

If \(x∈X\), denote by \([x]\) its equivalence class.

\(R\) is an equivalence relation if \(xℛx ; xℛy⇒yℛx ; xℛy, yℛz⇒xℛz\)

**Remark. **Giving a partition of a set \(X\) is equivalent to defining an equivalence relation on \(X\).

Let \((X, 𝒯)\) be a topological space and let \(ℛ⊂X×X\) be an equivalence relation on \(X\).

**Definition. **The set of equivalence classes of ℛ is denoted by \(X/ℛ\) and is called the **quotient space** of \(X\) with respect to ℛ.

The map \(p:X→X/ℛ\) is called the **collapsing map**.

**Proposition. **Let \((X, 𝒯)\) be a topological space and ℛ be an equivalence relation on \(X\). The family \(\tilde{𝒯}\) of sets \(\tilde{U}⊂X/ℛ\) such that \(p^{-1}(\tilde{U})∈𝒯\) is a topology on \(X/ℛ\) called the **quotient topology**.

**Proof. **(T1) \(p^{-1}(∅)=∅∈𝒯, p^{-1}(X/ℛ)=X∈𝒯\). So \(∅, X/ℛ∈\tilde{𝒯}\).

(T2) \(p^{-1}(\tilde{U}∩\tilde{V})=p^{-1}(\tilde{U})∩p^{-1}(\tilde{V})∈𝒯⇒\tilde{U}∩\tilde{V}∈\tilde{𝒯}\)

(T3) \(p^{-1}\left( \bigcup_{i∈I} \widetilde{U_i} \right)=\bigcup_{i∈I} p^{-1}(\widetilde{U_i})∈𝒯⇒\bigcup_{i∈I} \widetilde{U_i}∈\tilde{𝒯}\) ◻

**Remark. **By definition of quotient topology, \(p:X→X/ℛ\) is continuous.

**Proposition. **\(X\) is a topological space. ℛ is an equivalence relation. \(\tilde{V}⊂X/ℛ\) is closed in \(X/ℛ\) iff \(p^{-1}(\tilde{V})\) is closed in \(X\).

**Proof. **\(p^{-1}((X/ℛ)∖\tilde V)=X∖p^{-1}(\tilde{V})\)

\(\tilde{V}\) is closed\(⇔(X/ℛ)∖\tilde{V}\) is open\(⇔X∖p^{-1}(\tilde{V})\) is open\(⇔p^{-1}(\tilde{V})\) is closed. ◻

**Example.**

- \([0,1]∪[1, 2]/ℛ ≅[0, 2]\). Let \(1ℛ2\) and \(xℛx∀x\)
- \(X=[0,1], 0∼1, X/\mmlToken{mi}∼≅S^1\)

**Proposition 1. **Let \(X, Z\) be topological spaces and let ℛ be an equivalence relation of \(X\). Let \(g:X/ℛ→Z\) be a function. Then \(g\) is continuous iff \(g∘p:X→Z\) is continuous.

**Proof. **(\(⇒\)) \(g\) is continuous, \(p\) is continuous\(⇒g∘p\) is continuous

(\(⇐\)) Assume \(g∘p\) is continuous. Let \(U⊂Z\) be open.

\(p^{-1}(g^{-1}(U))=(g∘p)^{-1}(U)⇒p^{-1}(g^{-1}(U))\) is open

By definition of quotient space, \(g^{-1}(U)\) open\(⇒g\) is continuous ◻

Back to example \([0,1]/\mmlToken{mi}∼≅S^1\)

Define \(g:[0,1]/\mmlToken{mi}∼→S^1\) by \(g ([t])=e^{2πit}\)

\(0∼1\) but \(g (0)=e^{2πi 0}=e^{2πi 1}=g (1)\) so well-defined

\(g∘p:[0,1]→S^1\)

\(g∘p (t)=e^{2πit}\) which is continuous so by Prop 1 \(g\) is continuous.

\(g\) is 1-1, onto

\([0,1]\) is compact, so its image \(p ([0,1])=[0,1]/\mmlToken{mi}∼\) is compact.

\(S^1⊂ℝ^2\) is Hausdorff\(⇒S^1\) is Hausdorff

Recall: If \(f:X→Y\) is bijective, continuous, \(X\) is compact, \(Y\) is Hausdorff, then \(f\) is homeomorphism.

So \([0,1]/\mmlToken{mi}∼≅S^1\).

**Proposition 2. **Let \(f:X→Y\) be a surjective continuous map betwen topological spaces. Let ℛ be an equivalence relation on \(X\) defined by the partition \(\{f^{-1}(y):y∈Y\}\)

If \(X\) is compact, \(Y\) is Hausdorff, then \(X/ℛ, Y\) are homeomorphic.

**Proof. **Define \(g:X/ℛ→Y\) by \(g ([x])=f (x)\)

- \(g\) is well-defined by definition of ℛ

\(x∼y⇒f (x)=f (y)\) - \(g\) is onto since \(f\) is onto
- \(g\) is 1-1 since \(g ([x])=g ([y])⇒y∈f^{-1}(x)⇒x∼y\)
- \(X/ℛ= p (X)⇒X/ℛ\) is compact
- \(g ([x])=g (p (x))=f (x)\) is continuous\(⇒g\) is continuous by Prop 1

\(Y\) is Hausdorff, \(g:X/ℛ→Y\) is continuous bijection, so \(X/ℛ ≅Y\). ◻