Lecture 6

Econ 2001

2015 August 17

Lecture 6 Outline

1 Compactness2 Correspondences3 Continuity

In the first half, we talk about sets that are particularly useful, while in the secondhalf, we look at point to set mappings.

CompactnessDefinitionGiven a metric space (X , d), a collection of sets

U = {Uλ : λ ∈ Λ}is an open cover of A if Uλ is open for all λ ∈ Λ and

∪λ∈ΛUλ ⊃ A

Notice that Λ may be finite, countably infinite, or uncountable.

DefinitionGiven a metric space (X , d), a set A is compact if every open cover of A containsa finite subcover of A. In other words, if {Uλ : λ ∈ Λ} is an open cover of A, thereexist n ∈ N and λ1, · · · , λn ∈ Λ such that

A ⊂ Uλ1 ∪ · · · ∪ Uλn

The definition does not say “A has a finite open cover”(that would be silly...).

It requires that for any arbitrary open cover one can find a finite subcover ofthat given open cover.

Compactnes: Example

(0, 1] is not compactTo see this, let

U =

{Um =

(1m, 2)

: m ∈ N}

This is an open cover of (0, 1] since

∪m∈NUm = (0, 2) ⊃ (0, 1]

For (0, 1] to be compact we need to find a finite subcover of U .Take any finite collection {Um1 , . . . ,Umn} of elements U , and let

m = max{m1, . . . ,mn}Then

∪ni=1Umi = Um =

(1m, 2)6⊇ (0, 1]

So (0, 1] is not compact.

What about [0, 1]? This argument doesn’t work, so maybe that is compact...

Compactnes: Example

[0,∞) is closed but not compactTo see that [0,∞) is not compact, consider the open cover

U = {Um = (−1,m) : m ∈ N}

Compact requires that U must have a finite subcover.Take any finite subset of U

{Um1 , . . . ,Umn}As before, let

m = max{m1, . . . ,mn}Then

Um1 ∪ · · · ∪ Umn = Um = (−1,m) 6⊇ [0,∞)

Closed vs. CompactTheoremEvery closed subset A of a compact metric space (X , d) is compact.

Proof.Let {Uλ : λ ∈ Λ} be an open cover of A.

To use the compactness of X , we need an open cover of X .

Claim: U ′λ = Uλ ∪ (X \ A) is an open cover of X .

Since A is closed, X \ A is open; since Uλ is open, so is U ′λ.For any x ∈ X , either x ∈ A or x ∈ X \ A.

If x ∈ A, ∃λ ∈ Λ s.t. x ∈ Uλ ⊂ U ′λ. (since {Uλ : λ ∈ Λ} is an open cover of A.)If x ∈ X \ A, then ∀λ ∈ Λ, x ∈ U ′λ. (by the definition of U ′λ.)

Either way, x ∈ U ′λ and X ⊂ ∪λ∈ΛU ′λ; thus {U ′λ : λ ∈ Λ} is an open cover of X .

Since X is compact, there is a finite subcover of {U ′λ : λ ∈ Λ}∃λ1, . . . , λn ∈ Λ such that X ⊂ U ′λ1 ∪ · · · ∪ U

′λn

Then:a ∈ A⇒ a ∈ X ⇒ a ∈ U ′λi for some i ⇒ a ∈ Uλi ∪ (X \ A)⇒ a ∈ Uλi so

A ⊂ Uλ1 ∪ · · · ∪ UλnThus A is compact.

Compactnesss and ClosednessAlthough closed 6⇒ compact (see earlier example), the converse is true:

TheoremIf A is a compact subset of the metric space (X , d), then A is closed.

Proof.By contradiction: suppose A is not closed. Then X \ A is not open; so there isx ∈ X \ A such that, for every ε > 0, A ∩ Bε(x) 6= ∅, and hence A ∩ Bε[x ] 6= ∅.

For n ∈ N, let Un = X \ B 1n[x ]

Each Un is open, and since x 6∈ A: ∪n∈NUn = X \ {x} ⊃ A .

Therefore, {Un : n ∈ N} is an open cover of A.Since A is compact, there is a finite subcover {Un1 , . . . ,Unk }.Let n = max{n1, . . . , nk}. Then

Un = X \ B 1n[x ] ⊃ X \ B 1

nj[x ]︸ ︷︷ ︸

j=1,...,k

= ∪kj=1Unj ⊃ A

But A ∩ B 1n[x ] 6= ∅, so A 6⊆ X \ B 1

n[x ] = Un , a contradiction.

Thus A is closed.

Sequential Compactness

DefinitionA set A in a metric space (X , d) is sequentially compact if every sequence ofelements of A contains a convergent subsequence whose limit lies in A.

This is an alternative characterization of compactness.

TheoremA set A in a metric space (X , d) is compact if and only if it is sequentially compact.

Proof? Not obvious.

Easy Heine-Borel Theorem for R

TheoremIf A ⊂ E1, then A is compact if and only if A is closed and bounded.

Proof.Let A be a closed, bounded subset of R.Then A ⊂ [a, b] for some interval [a, b].

Let {xn} be a sequence of elements of [a, b].

By the Bolzano-Weierstrass Theorem, {xn} contains a convergentsubsequence; let x ∈ R denote the limit.Since [a, b] is closed, x ∈ [a, b].

Thus, [a, b] is sequentially compact, hence compact. Since A is a closedsubset of [a, b] it is also compact.

Conversely, if A is compact, A is closed and bounded by a previous result.

Heine-Borel Theorem

Theorem (Heine-Borel)If A ⊆ En , then A is compact if and only if A is closed and bounded.

ExampleA closed interval in Rn is defined as:

[a, b] = {x ∈ Rn : ai ≤ xi ≤ bi for each i = 1, . . . , n}[a, b] is compact in En for any a, b ∈ Rn .

Continuous Functions and Compact SetsContinuous Images of Compact Sets Are Compact

TheoremLet (X , d) and (Y , ρ) be metric spaces. If f : X → Y is continuous and C is acompact subset of (X , d), then f (C ) is compact in (Y , ρ).

Proof.Problem Set 6.

Continuous functions of compact sets are uniformly continuous.

TheoremLet (X , d) and (Y , ρ) be metric spaces, C a compact subset of X , and f : C → Ycontinuous. Then f is uniformly continuous on C.

Proof.Problem Set 6.

Extreme Value Theorem (Again!)

Corollary (Extreme Value Theorem)Let C be a compact set in a metric space (X , d), and suppose f : C → R iscontinuous. Then f is bounded on C and attains its minimum and maximum on C.

Proof.f (C ) is compact by a previous Theorem, hence it is closed and bounded.

Let M = sup f (C ); w know M <∞.Then there exists ym ∈ f (C ) such that

M − 1m≤ ym ≤ M

so M is a limit point of f (C ).

Since f (C ) is closed, M ∈ f (C ), i.e. there exists c ∈ C such thatf (c) = M = sup f (C ), so f attains its maximum at c .

The proof for the minimum is similar (do it).

Theorem of the Maximum: a PreludeA major interest in economics is constrained optimzation. In particular, wewould like to know how the maximum value and the maximizing vectors in aconstrained maximization problem depend on some exogenous parameters.

Let f : A× X → R, where A ⊂ Rm and X ⊂ Rn .For each a ∈ A, let ϕ(a) be a non-empty subset of X .

A maximization problem is defined as

maxb∈ϕ(a)

f (a, x).

For each a ∈ A, leth(a) = max

b∈ϕ(a)f (a, x) and µ(a) = {b ∈ ϕ(a) : h(a) = f (a, x)}

These are, respectively, the maximized value of the objective and the maximalchoice, as functions of the parameters a.

ϕ(a) is typically a set, and µ(a) is typically not unique (µ(a) ⊂ X ). These arepoint to set mappings.

The Theorem of the Maximum gives conditions for h and µ to be continuousin a.

But what does continuity mean for set valued mappings?

Correspondences

A set valued mapping takes each point in the domain to a set in the range.

DefinitionA correspondence from X to Y , denoted ϕ : X → 2Y , is a mapping from X to 2Y ;that is, ϕ(x) ⊂ Y for every x ∈ X .

This maps every point in X to a subset of Y .

Obviously, a function from X to Y is a correspondence that is single-valuedfor each x .

Let f : X → Y be a function. Define ϕ : X → 2Y by

ϕ(x) = {f (x)} for each x ∈ X

A function is a special case of a correspondence.

We typically deal with metric spaces (X , d) and (Y , d), where X ⊂ Rn andY ⊂ Rm , and d is the Euclidean metric.

Correspondences Are Commonplace in EconomicsThe Budget Set is a CorrespondenceA consumer uses her income to buy any combination of the L goods at given prices.

Prices are L non-negative numbers measuring how much each unit of eachgood costs. They are a vector in RL+.Possible consumption choices are described by L non negative quantities, onefor each good. This is also a vector in RL+.Given prices and income, the consumer can only choose combinations of thegoods that she can afford:

∑Ll=1 plxl ≤ I .

For given prices and income, the affordable consumption bundles for a set{x ∈ RL+ : p · x =

L∑l=1

plxl ≤ I}

The correspondence B : RL+ × R++ → RL+ defined as follows:

B(p, I ) =

{x ∈ RL+ : p · x =

L∑l=1

plxl ≤ I}⊂ RL+

describes all affordable consumption bundles (elements of RL+) for differentprice vectors and income levels.

Correspondences Are Useful In Economics

A Consumer’s Demand is a Correspondence

The consumer ranks consumption bundles using a utility function u : RL+ → R.

Define x∗ : RL++ × R++ → 2RL+ by

x∗(p, I ) = arg maxx∈B(p,I )

u(x)

This is a consumer’s demand and can be multi-valued (and thus acorrespondence).

A typical question an economist likes to answer is: what happens to demandwhen prices and/or income change? That is, how does x∗ change as p and Ichange?

You can answer this questions if you know what the correspondence x∗(p, I )looks like.

Why assume that income is strictly positive? What if income is zero?

Why assume that all prices are strictly positive? What if some prices are zero?

Definitions for Correspondences

Since correspondences are sets, one can talk about closedness and/orcompactness of these sets.

DefinitionA correspondence ϕ : X → 2Y is closed-valued if ϕ(x) is a closed subset of Y forall x .

DefinitionA correspondence ϕ : X → 2Y is compact-valued if ϕ(x) is compact for all x .

The graph of a correspondence is defined similarly to the graph of a function.

DefinitionThe graph of a correspondence ϕ : X → 2Y is the set

Gϕ = {(x , y) ∈ X × Y : y ∈ ϕ(x)}

Continuity for Correspondences

We will see three different notions of continuity for correspondences, eachmotivated by continuity for functions.

A function f : R→ R may be discontinuous at a point x0 because it “jumpsdown:”

∃xn → x0 such that f (x0) < lim inf f (xn)

or it “jumps up”

∃xn → x0 such that f (x0) > lim sup f (xn)

In either case, it does not matter whether the sequence xn approaches x0 fromthe left or the right, the function is not continuous at x0.

For sets, however, this matters and makes for different possible views ofcontinuity. Hence all the different definitions.

Continuity for CorrespondencesHow can a set “jump”at the limit x0? Maybe the set suddenly gets smaller.

There is a sequence xn → x0, with image ϕ(x0), but as n→∞ the pointsyn ∈ ϕ(xn) are far from every point of ϕ(x0).

Continuity for CorrespondencesA second way a set can “jump”at the limit is to suddenly get larger.

There is a point y in ϕ(x0) and a sequence xn → x0 such that y is far fromevery point of ϕ(xn) as n→∞.

Upper Hemicontinuity

DefinitionLet X ⊂ Rn , Y ⊂ Rm , and ϕ : X → 2Y .

ϕ is upper hemicontinuous (uhc) at x0 ∈ X if,

for every open set V ⊃ ϕ(x0)

there is an open set U with x0 ∈ U

suchthat

ϕ(x) ⊂ Vfor every x ∈ U ∩ X

ϕ is upper hemicontinuous if it is upper hemicontinuous at every x ∈ X .

Draw a Picture

In other words, for every x0 ∈ X and every ε > 0, there is a δ > 0 such thatϕ(x) ⊂ Bε(ϕ(x0)) when ‖x − x0‖ < δ.

Lower Hemicontinuity

DefinitionLet X ⊂ Rn , Y ⊂ Rm , and ϕ : X → 2Y .

ϕ is lower hemicontinuous (lhc) at x0 ∈ X if,

for every open set V such that ϕ(x0) ∩ V 6= ∅

there is an open set U with x0 ∈ U

suchthat

ϕ(x) ∩ V 6= ∅for every x ∈ U ∩ X

ϕ is lower hemicontinuous if it is lower hemicontinuous at every x ∈ X .

Draw a Picture

Continuity for Correspondences

A correspondence is continuous if it is both upper and lower hemicontinuous.

DefinitionLet X ⊂ Rn , Y ⊂ Rm , and ϕ : X → 2Y .

ϕ is continuous at x0 ∈ X if it is both upper hemicontinuous and lowerhemicontinuous at x0.

ϕ is continuous if it is continuous at every x ∈ X .

Hemicontinuity and FunctionsIf the correspondence is a function, upper hemi-continuity and continuity coincide.

TheoremLet X ⊂ Rn , Y ⊂ Rm and f : X → Y . Let ϕ : X → 2Y be given by ϕ(x) = {f (x)}for all x ∈ X. Then ϕ(x) is uhc if and only if f is continuous.

Proof.Suppose ϕ is uhc. Fix V open in Y .

Then

f −1(V ) = {x ∈ X : f (x) ∈ V }= {x ∈ X : ϕ(x) ⊂ V }

Thus, f is continuous if and only if f −1(V ) is open in X for each open V in Y .

Hence, if and only if {x ∈ X : ϕ(x) ⊂ V } is open in X for each open V in Y .

Therefore, if and only if ϕ is uhc (think through why this holds).

There is a similar statement for lower hemicontinuity. Prove it.

Hemicontinuity Notions Are Not NestedIn the picture, the correspondece violates uhc at x∗ (but not lhc) and it violates lhcbetween x∗ and x0 (but not uhc).

uhc at x0 ∈ X : ∀ open set V ⊇ ϕ(x0 ), ∃ an open set U with x0 ∈ U s.t. ϕ(x) ⊆ V , ∀x ∈ U ∩ X

lhc at x0 ∈ X : ∀ open set V s.t. ϕ(x0 ) ∩ V 6= ∅, ∃ an open set U with x0 ∈ U s.t. ϕ(x) ∩ V 6= ∅, ∀x ∈ U ∩ X

Continuity and the Graph

Recall that a function f : Rn → R is continuous if and only if whenever xn → x ,f (xn)→ f (x). We can translate this into a statement about its graph.

The graph of function f : X → Y is the setGf = {(x , y) ∈ X × Y : y = f (x)}.Suppose {(xn , yn)} ⊂ Gf and (xn , yn)→ (x , y). Since f is a function,

(xn , yn) ∈ Gf ⇔ yn = f (xn).

So

f is continuous ⇒ y = lim yn = lim f (xn) = f (x)

⇒ (x , y) ∈ GfSo if f is continuous then each convergent sequence {(xn , yn)} in Gfconverges to a point (x , y) also in Gf .

Thus, if f is continuous then Gf is a closed set.

Continuity for Correspondences: Closed Graph

Remember, the graph of a correspondence ϕ : X → 2Y is the set

Gϕ = {(x , y) ∈ X × Y : y ∈ ϕ(x)}

DefinitionLet X ⊆ Rn , Y ⊆ Rm . A correspondence ϕ : X → 2Y has closed graph if its graphis a closed subset of X × Y .

REMARKThe condition that Gϕ is closed can be stated using sequences.

For any sequences {xn} ⊆ X and {yn} ⊆ Y such that

xn → x ∈ Xyn → y ∈ Y

andyn ∈ ϕ(xn) for each n

⇒ y ∈ ϕ(x)

Closed Graph: Example

Closed Graph but Not Upper Hemicontinuous

Consider the correspondence ϕ(x) =

{ {1x

}if x ∈ (0, 1]

{0} if x = 0

Let V = (−0.1, 0.1). Then ϕ(0) = {0} ⊂ V , but no matter how close x is to0, ϕ(x) =

{1x

}6⊆ V

So ϕ is not uhc at 0. However, note that ϕ has closed graph.

Closed Graph: Examples

Closed Graph and Upper Hemicontinuous

Consider the correspondence ϕ(x) =

{ {1x

}if x ∈ (0, 1]

R+ if x = 0

ϕ(0) = [0,∞), and ϕ(x) ⊂ ϕ(0) for every x ∈ [0, 1].

So if V ⊃ ϕ(0) then V ⊃ ϕ(x) for all x .

Thus, ϕ is uhc, and has closed graph.

Hemicontinuity and Closed GraphA closed graph correspondence need not be uhc (see above), and an uhccorrespondence need not have closed graph, or even have closed values.A closed graph correspondence need not be lhc.

Continuity in Special Cases

The three continuity concepts are connected in special cases.

A closed-valued upper hemicontinuous correspondence must have closed graph.For a closed-valued correspondence with a compact range, upperhemicontinuity is equivalent to closed graph.

TheoremSuppose X ⊂ Rn , Y ⊂ Rm , and ϕ : X → 2Y .

(i) If ϕ is closed-valued and uhc, then ϕ has closed graph.

(ii) If ϕ has closed graph and

there arean open set W with x0 ∈W

anda compact set Z such that x ∈W ∩ X ⇒ ϕ(x) ⊂ Z

⇒ ϕ is uhc at x0

(iii) If Y is compact, then ϕ has closed graph ⇐⇒ ϕ is closed-valued and uhc.

(iii) Follows from (i) and (ii) so there is no need to prove it.