Topology and Continuous functions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 5 March 2011

Topology

A topological space is a set $X$ with a specification of the open subsets of $X$ where it is required that

(a)

\emptyset

is open and

X

is open,

(b) Unions of open sets are open,

In other words, a topology on

X

is a set

𝒯

of subsets of

X

such that

(a)

\emptyset \in 𝒯

and

X \in 𝒯

(b) If

𝒮 \subseteq 𝒯

then

(⋃_{U \in 𝒮} U) \in 𝒯

n \in ℤ_{\geq 0}

and

U_{1} U_{2} \dots U_{n} \in 𝒯

then

U_{1} \cap U_{2} \cap \dots \cap U_{n} \in 𝒯 .

A topological space is a set $X$ with a topology $𝒯$ on $X$ .

Let $𝒯$ be a topology on $X$ . An open set is a set in $𝒯$ .

A closed set is a subset $E$ of $X$ such that the complement $E^{c}$ of $E$ is open.

Let $X$ be a topological space. A subspace of $X$ is a subset $E$ of $X$ with the topology given by making the open sets be the sets $ι^{- 1} (V) such that V is an open subset of X,$ where $ι : E \to X$ is the inclusion.

A connected set is a subset $E \subseteq X$ such that there do not exist open sets $A$ and $B$ with $A \cap E \neq \emptyset, B \cap E \neq \emptyset, A \cup B \supseteq E, and (A \cap B) \cap E = \emptyset .$ A compact set is a subset $E \subseteq X$ such that every open cover of $E$ contains a finite subcover. More precisely, a compact set is a subset $E \subseteq X$ such that

𝒮 \subseteq 𝒯

and

⋃_{U \in 𝒮} U \supseteq E

then there exists

n \in ℤ_{> 0}

and

U_{1}, \dots, U_{n} \in 𝒮

such that

(U_{1} \cup U_{2} \cup \dots \cup U_{n}) \supseteq E .

Continuous Functions

Continuous functions are for comparing topological spaces.

Let $X$ and $Y$ be topological spaces. A function $f : X \to Y$ is continuous if it satisfies the condition

V

is an open subset of

Y

then

f^{- 1} (V)

is an open subset of

X

Let $X$ and $Y$ be topological spaces. Let $a \in X$ . A function $f : X \to Y$ is continuous at $a$ if it satisfies the condition

V

is a neighborhood of

a

Y

then

f^{- 1} (V)

is a neighborhood of

f^{- 1} (a)

X

Here

f^{- 1} (V) = {x \in X | f (x) \in V} .

Let $X$ and $Y$ be topological spaces and let $a \in X$ . A function

f : X \to Y

is continuous at

a

if and only if

lim_{x \to a} f (x) = f (a).

Let $X$ and $Y$ be topological spaces. An isomorphism, or homeomorphism, is a continuous function $f : X \to Y$ such that the inverse function $f^{- 1} : Y \to X$ exists and is continuous.

Let $f : X \to Y$ be a continuous function and let $E \subseteq X$ .

(a) If

E

is connected then

f (E)

is connected.

(b) If

E

is compact then

f (E)

is compact.

		Proof of (a):
	Proof by contradiction. Assume $f (E)$ is not connected. Let $A$ and $B$ be open in $f (E)$ such that $A \neq \emptyset$ and $B \neq \emptyset$ and $A \cup B \supseteq f (E)$ and $A \cap B = \emptyset$ . Then let $C = f^{- 1} (A)$ and $D = f^{- 1} (B)$ . Then $C \cup D = f^{- 1} (A) \cup f^{- 1} (B) = f^{- 1} (A \cup B) \supseteq f^{- 1} (f (E)) \supseteq E$ , and $C \cap D = f^{- 1} (A) \cap f^{- 1} (B) = f^{- 1} (A \cap B) = f^{- 1} (\emptyset) = \emptyset$ . Now $C \neq \emptyset$ since $A \neq \emptyset$ and $A \subseteq f (E)$ , and $D \neq \emptyset$ since $B \neq \emptyset$ and $B \subseteq f (E)$ . So $E$ is not connected. This is a contradiction. So $f (E)$ is connected. $□$

		Proof of (b):
	Assume $f : X \to Y$ is continuous and $E$ is compact. To show: If $𝒮$ is an open cover of $f (E)$ then it has a finite subcover. To show: If $𝒴$ is the topology on $Y$ , $𝒮 \subseteq 𝒴$ , and $(⋃_{V \in 𝒮} V) \supseteq f (E)$ then there exists $n \in ℤ_{> 0}$ and $V_{1}, \dots, V_{n} \in 𝒮$ such that $V_{1} \cup \dots \cup V_{n} \supseteq f (E)$ . Assume $𝒴$ is the topology on $Y$ and $𝒮 \subseteq 𝒴$ and $(⋃_{V \in 𝒮} V) \supseteq f (E)$ . Then $(⋃_{V \in 𝒮} f^{- 1} (V)) \supseteq E$ . Let $𝒯 = {f^{- 1} (V) \| V \in 𝒮}$ . Since $f$ is continuous, $f^{- 1} (V)$ is open. Thus $𝒯 \subseteq 𝒳$ where $𝒳$ is the topology of $X$ , and also $(⋃_{f^{- 1} (V) \in 𝒯} f^{- 1} (V)) \supseteq E$ . Since $E$ is compact there exists $n \in ℤ_{> 0}$ and $f^{- 1} (V_{1}), \dots, f^{- 1} (V_{n}) \in 𝒯$ such that $(f^{- 1} (V_{1}) \cup \dots \cup f^{- 1} (V_{n})) \supseteq E$ . So $(V_{1} \cup \dots \cup V_{n}) \supseteq f (E)$ . Thus $𝒮$ contains a finite subcover of $f (E)$ . $□$

Examples

(1) Let $X$ be a set. The discrete topology on $X$ is the topology such that every subset of $X$ is open.

(2) A metric space is a set $X$ with a function $d : X \times X \to ℝ_{\geq 0}$ such that

(a) If

x \in X

then

d (x, x) = 0

(b) If

x, y \in X

and

d (x, y) = 0

, then

x = y

x, y, z \in X

then

d (x, z) \leq d (x, y) + d (y, z)

Let $X$ be a metric space. Let $x \in X$ and let $ε \in ℝ_{> 0}$ . The ball of radius $ε$ at $x$ is the set

B_{ε} (x) = {p \in X | d (x, y) \leq ε}

Let

X

be a metric space. The metric space topology on

X

is the topology generated by the sets

B_{ε} (x),

for

x \in X

and

ε \in ℝ_{> 0}

Homework

$f^{- 1} (C \cup D) = f^{- 1} (C) \cup f^{- 1} (D)$ .
$f^{- 1} (C \cap D) = f^{- 1} (C) \cap f^{- 1} (D)$ .
$f^{- 1} (f (E)) \supseteq E$ .
Give an example where $f^{- 1} (f (E)) \neq E$ .
If $E$ is compact then $E$ is closed.

Notes and References

This summary of the theory of constructible functions is part of joint work with A. Ghitza and S. Kannan on the relationship between MV-cycles and the Borel-Weil-Bott theorem. This presentation follows [GLS, Section 4.1].

References

[GLS] C. Geiss, B. Leclerc and J. Schröer, Semicanonical bases and preprojective algebras, Ann. Sc. École Norm. Sup. 38 (2005), 193-253. (2003), 567-588, arXiv:math/0402448, MR2144987.

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