Metric and Hilbert Spaces

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

Last updated: 4 November 2014

Lecture 7: Connectedness and connected components

Homework questions

(1) Let X be a set. Show that the discrete topology on X is a topology on X.
(2) Let (X,d) be a metric space. Show that the metric space topology on X is a topology on X.
(3) Let (X,𝒯) be a topological space and let YX. Show that the subspace topology on Y is a topology on Y.
(4) Let (X,𝒯X) and (Y,𝒯Y) be topological spaces. Show that the product topology on X×Y is a topology on X×Y.

One could do these, one by one, directly, using proof machine. Alternatively, one can develop a few helpful definitions that, more or less, do them all in one fell swoop.

Let X be a topological space with topology 𝒯.

A base of the topology on X is a collection B𝒯 such that if U𝒯 then there exists 𝒮B such that U=A𝒮A.

Let xX. A fundamental system of neighbourhoods of x is a set 𝒮𝒩(x) such that if N𝒩(x) then there exists W𝒮 such that WN.

HW 1 Show that B is a base of 𝒯 if and only if B satisfies if xX then {VB|xV} is a fundamental system of neighbourhoods of x.

HW 2 Let (X,d) be a metric space. Show that the set of open balls is a base of the metric space topology on X by showing that it satisfies the condition in HW 1.

HW 3 Let (X,𝒯X) and (Y,𝒯Y) be topological spaces. Show that B={U×V|U𝒯XandV𝒯Y} is a base of the product topology on X×Y by showing that it satisfies the condition in HW 1.

Let (X,𝒯) be a topological space.

A connected set is a subset EX such that there do not exist open sets A and B (A,B𝒯) with AEand BEand ABEand (AB)E=. Perhaps it is better to think of E with the subspace topology 𝒯E. Then E is connected if there do not exist U and V open in E (U,V𝒯E) such that UandV andUV=E andUV=.

Let (X,𝒯X) and (Y,𝒯Y) be topological spaces. Let f:XY be a function.

The function f:XY is continuous if f satisfies: if V𝒯Y then f-1(V)𝒯X.

Recall that, by definition, f-1(V)= {xX|f(x)V}.

Recall that, by definition, a function f:XY is a subset ΓX×Y such that if xX then there exists a unique yY such that (x,y)Γ. Use the notation f(x) so that Γ={(x,f(x))|xX}.

Let f:XY be a continuous function. Let EX. If E is connected
then f(E) is connected.

Proof.

Assume E is connected.
To show: f(E) is connected.
Proof by contradiction.
Assume f(E) is not connected.
Let A and B be open in f(E) such that Aand Band ABf(E) andAB=. Let C=f-1(A) and D=f-1(B).
Then CD=f-1(A) f-1(B)= f-1(AB) f-1(f(E)) E and CD=f-1(A) f-1(B)= f-1(AB)= f-1()= and CsinceA andAf(E), DsinceB andBf(E). So E is not connected. This is a contradiction.
So f(E) is connected.

Notes and References

These are a typed copy of Lecture 7 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 7, 2014.

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