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 18: Connectedness, compactness and the Mean value theorem

Let X,Y be topological spaces and let f:XY be a continuous function. Let EX.

(a) If E is cover compact then f(E) is cover compact.
(b) If E is connected then f(E) is connected.

Let f:X be continuous function where X is a compact metric space. Then f attains a maximum and minimum value, i.e. there exist aXsuch that f(a)=inf {f(x)|xX} and bXsuch that f(b)=sup {f(x)|xX}.

Let A.

(a) A is connected if and only if A is an interval.
(b) A is compact and connected then A is a closed bounded interval.

(Rolle's theorem) f:[a,b] with f(a)=f(b).

(Mean value theorem) f:[a,b]. There exists c[a,b] with f(c)= f(b)-f(a)b-a.

Sketches of proofs.

(a) cover compact sequentially compact
To show: not sequentially compact not cover compact.
Let a1,a2, be a sequence in A with no cluster point.
For each aA let εa>0 be such that B(a,εa){a1,a2,} is finite.
Then {B(a,εa)|aA} is an open cover of A with no finite subcover.
So A is not cover compact.
(b) sequentially compact cover compact
To show: not cover compact not sequentially compact.
Let 𝒮 be a cover of A with no finite subcover.
Let a1A and S1𝒮 with a1S1.
Let a2A\S1 and S2𝒮 with a2S2.
Let a3A\(S1S2) and S3𝒮 with a3S3.

Then a1,a2,a3, is a sequence in A with no cluster point.
(A cluster point a would have an S𝒮 with aS, and so S is a neighbourhood of a and would contain all but a finite number of ai and so S1S2SNS would be a finite cover??)

Notes and References

These are a typed copy of Lecture 18 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 27, 2014.

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