Real Analysis

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

Last update: 19 July 2014

Lecture 29

Topology

Goal: Understand and prove the Intermediate value theorem and The max-min theorem.

(Intermediate Value Theorem) Let f:[a,b]→ℝ be a continuous function. If x∈ℝ and x is between f(a) and f(b) then there exists c∈[a,b] with f(c)=x.

(Max-Min theorem) Let f:[a,b]→ℝ be a continuous function. Then there exists m,M∈ℝ such that f([a,b]) =[m,M].

Let X be a set.

A topology on X is a collection 𝒯 of subsets of X such that

(a) ∅∈𝒯 and X∈𝒯,
(b) if 𝒮⊆𝒯 then ⋃u∈𝒮u∈𝒯,
(c) if 1,…,n∈ℤ>0 and U1,…,Un∈𝒯 then U1∩⋯∩Un∈𝒯.

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

Let X be a topological space with topology 𝒯. Let E be a subset of X.

The set E is open is E∈𝒯.

Let X={a,b,c,d}. 𝒯 = { ∅, {a}, {b,c}, {a,b,c}, {d}, {b,c,d}, {a,b,c,d} } is not a topology on X. 𝒯 = { ∅, {a}, {d}, {b,c}, {a,d}, {a,b,c}, {b,c,d}, {a,b,c,d} } is a topology on X.

Let X=ℝn. 𝒯={open subsets E⊆ℝn} is a topology on ℝn (a subset E⊆ℝn is open if E is a union of ε-balls).

Let X be a topological space with topology 𝒯.

Let E be a subset of X.

The set E is closed if Ec is open.

Recall: Ec={x∈X | x∉E}.

The set E is connected if there do not exist open sets U1,U2 such that U1∪U2⊇Eand (U1∩E)∩ (U2∩E)=∅.

The set E is compact is E satisfies: If 𝒮⊆𝒯 and (⋃u∈𝒮u)⊇E then there exists n∈ℤ>0 and U1,…,Un∈𝒮 such that (U1∪U2∪⋯∪Un)⊇E. In English: E is compact if every open cover of E has a finite subcover.

What does this have to do with the Max-min theorem?

Let X=ℝ with the standard topology. Let E⊆ℝ. Then E is compact and connected
if and only if
there exist m,M∈ℝ such that E=[m,M].

Continuous functions for topological spaces

Let X and Y be sets. Let f:X→Y be a function. Let E be a subset of X. The image of E is f(E)= {f(x) | x∈E}. X Y E f(E) Let F be a subset of Y. the inverse image of F is f-1(F)= {x∈X | f(x)∈F}. X Y F f-1(F)

Warning: Despite the confusing notation, f-1(F) has nothing to do with the inverse function to f.

Let X be a topological space with topology 𝒯. Let Y be a topological space with topology 𝒵. Let f:X→Y be a function. The function f:X→Y is continuous if f satisfies: If V∈𝒵 then f-1(V)∈𝒯. In English: Inverse images of open sets are open.

Let a,b∈ℝ with a<b. Let X=[a,b] with the topology given by E⊆X is open if E is a union of ε-balls. Let Y=ℝ with the standard topology.

Let f:[a,b]→ℝ. The function f is continuous (as a function between topological spaces) if and only if f satisfies if c∈[a,b] then limx→cf(x)=f(c).

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100517Lect29.pdf and was given on 17 May 2010.

page history