Properties of the set of continuous functions from a compact Hausdorff space to a topological field - Part I

Properties of the set of continuous functions from acompact Hausdorff space to a topological field

Pawel [email protected]

Abstract Algebra Seminar, Knoxville, TN

December 12, 2016

Pawel Grzegrzolka [email protected] Properties of the set of continuous functions from a compact Hausdorff space to a topological fieldDecember 12, 2016 1 / 1

Throughout this talk we will use the following definitions:

Definition

A topological space F which is also a field is called a topological field ifthe maps (a, b)→ a + b, (a, b)→ ab are continuous maps from F × F(equipped with the product topology) to F , and the map a→ a−1 is acontinuous map from F \ {0} to F \ {0}.

Definition

Let X be a compact, Hausdorff space. Then C (X ,F ) denotes the set ofall continuous functions from X to F .


Throughout this talk we will use the following definitions:

Definition

A topological space F which is also a field is called a topological field ifthe maps (a, b)→ a + b, (a, b)→ ab are continuous maps from F × F(equipped with the product topology) to F , and the map a→ a−1 is acontinuous map from F \ {0} to F \ {0}.

Definition

Let X be a compact, Hausdorff space. Then C (X ,F ) denotes the set ofall continuous functions from X to F .


Definition

Let X be a compact, Hausdorff space and F a topological field. For eachy ∈ X , let

My := {f ∈ C (X ,F ) | f (y) = 0}.

Definition

Let X be a topological space. Then X is said to be normal if for each pairA,B of disjoint closed sets of X , there exists disjoint open sets containingA and B, respectively.


Definition

Let X be a compact, Hausdorff space and F a topological field. For eachy ∈ X , let

My := {f ∈ C (X ,F ) | f (y) = 0}.

Definition

Let X be a topological space. Then X is said to be normal if for each pairA,B of disjoint closed sets of X , there exists disjoint open sets containingA and B, respectively.


Throughout this talk, we will assume that X is a compact Hausdorff spaceand that F is a topological field.

Proposition

C (X ,F ) is a commutative ring with 1 6= 0 under the pointwise additionand multiplication of functions.


Throughout this talk, we will assume that X is a compact Hausdorff spaceand that F is a topological field.

Proposition

C (X ,F ) is a commutative ring with 1 6= 0 under the pointwise additionand multiplication of functions.


Proof.

Notice that C (X ,F ) is non-empty, since the zero map sending everyelement to 0 (by abuse of notation, we will denote this map 0), and theconstant map sending every element to 1 (by abuse of notation, we willdenote this map 1) are continuous functions from X to F .

Clearly, 0 6= 1.Also, for any f ∈ C (X ,F ) and any x ∈ X ,

(f + 0)(x) = f (x) + 0(x) = f (x),

(0 + f )(x) = 0(x) + f (x) = f (x),

and(f · 1)(x) = f (x) · 1(x) = f (x),

(1 · f )(x) = 1(x) · f (x) = f (x).

Thus, 0 is the additive identity element and 1 is the unity.


Proof.

Notice that C (X ,F ) is non-empty, since the zero map sending everyelement to 0 (by abuse of notation, we will denote this map 0), and theconstant map sending every element to 1 (by abuse of notation, we willdenote this map 1) are continuous functions from X to F . Clearly, 0 6= 1.

Also, for any f ∈ C (X ,F ) and any x ∈ X ,

(f + 0)(x) = f (x) + 0(x) = f (x),

(0 + f )(x) = 0(x) + f (x) = f (x),

and(f · 1)(x) = f (x) · 1(x) = f (x),

(1 · f )(x) = 1(x) · f (x) = f (x).



Proof.

Notice that C (X ,F ) is non-empty, since the zero map sending everyelement to 0 (by abuse of notation, we will denote this map 0), and theconstant map sending every element to 1 (by abuse of notation, we willdenote this map 1) are continuous functions from X to F . Clearly, 0 6= 1.Also, for any f ∈ C (X ,F ) and any x ∈ X ,

(f + 0)(x) = f (x) + 0(x) = f (x),

(0 + f )(x) = 0(x) + f (x) = f (x),

and(f · 1)(x) = f (x) · 1(x) = f (x),

(1 · f )(x) = 1(x) · f (x) = f (x).



Proof (Cont.)

Furthermore, notice that for any f , g ∈ C (X ,F ), f + g is a continuousfunction, being the composition of continuous functions, namely

X −→ F × F −→ Fx −→ f (x)× g(x) −→ f (x) + g(x),

where the first function is continuous by being continuous in each variable,and the second function is continuous by the definition of the topologicalfield. Thus, C (X ,F ) is closed under addition.

Similarly, f · g iscontinuous, since

X −→ F × F −→ Fx −→ f (x)× g(x) −→ f (x) · g(x),

is a composition of continuous functions. Thus, C (X ,F ) is closed undermultiplication.


Proof (Cont.)

Furthermore, notice that for any f , g ∈ C (X ,F ), f + g is a continuousfunction, being the composition of continuous functions, namely

X −→ F × F −→ Fx −→ f (x)× g(x) −→ f (x) + g(x),

where the first function is continuous by being continuous in each variable,and the second function is continuous by the definition of the topologicalfield. Thus, C (X ,F ) is closed under addition. Similarly, f · g iscontinuous, since

X −→ F × F −→ Fx −→ f (x)× g(x) −→ f (x) · g(x),

is a composition of continuous functions. Thus, C (X ,F ) is closed undermultiplication.


Proof (Cont.)

Since constant functions are continuous, for any f ∈ C (X ,F ),−f ∈ C (X ,F )

By associativity of addition and multiplication in F ,C (X ,F ) is associative under addition and multiplication. Since fieldmultiplication is distributive over field addition, multiplication in C (X ,F )is distributive over addition. Finally, since F is a field, its multiplication iscommutative, which implies that multiplication in C (X ,F ) iscommutative.


Proof (Cont.)

Since constant functions are continuous, for any f ∈ C (X ,F ),−f ∈ C (X ,F ) By associativity of addition and multiplication in F ,C (X ,F ) is associative under addition and multiplication.

Since fieldmultiplication is distributive over field addition, multiplication in C (X ,F )is distributive over addition. Finally, since F is a field, its multiplication iscommutative, which implies that multiplication in C (X ,F ) iscommutative.


Proof (Cont.)

Since constant functions are continuous, for any f ∈ C (X ,F ),−f ∈ C (X ,F ) By associativity of addition and multiplication in F ,C (X ,F ) is associative under addition and multiplication. Since fieldmultiplication is distributive over field addition, multiplication in C (X ,F )is distributive over addition.

Finally, since F is a field, its multiplication iscommutative, which implies that multiplication in C (X ,F ) iscommutative.


Proof (Cont.)

Since constant functions are continuous, for any f ∈ C (X ,F ),−f ∈ C (X ,F ) By associativity of addition and multiplication in F ,C (X ,F ) is associative under addition and multiplication. Since fieldmultiplication is distributive over field addition, multiplication in C (X ,F )is distributive over addition. Finally, since F is a field, its multiplication iscommutative, which implies that multiplication in C (X ,F ) iscommutative.


Proposition

My is a maximal ideal of C (X ,F ) for all y ∈ X .

Proof.

Let y ∈ X be arbitrary. Notice that My is non-empty, since 0 ∈ My . Nownotice that if f , g ∈ My and h ∈ C (X ,F ), then

(f − g)(y) = f (y)− g(y) = 0,

(h · f )(y) = h(y) · f (y) = h(y) · 0 = 0.

Thus, (f − g) and h · f are in My , implying that My is an ideal. Alsonotice that My 6= C (X ,F ), since 1 /∈ My .


Proposition


Proof.

Let y ∈ X be arbitrary. Notice that My is non-empty, since 0 ∈ My .

Nownotice that if f , g ∈ My and h ∈ C (X ,F ), then

(f − g)(y) = f (y)− g(y) = 0,

(h · f )(y) = h(y) · f (y) = h(y) · 0 = 0.



Proposition


Proof.


(f − g)(y) = f (y)− g(y) = 0,

(h · f )(y) = h(y) · f (y) = h(y) · 0 = 0.

Thus, (f − g) and h · f are in My , implying that My is an ideal.

Alsonotice that My 6= C (X ,F ), since 1 /∈ My .


Proposition


Proof.


(f − g)(y) = f (y)− g(y) = 0,

(h · f )(y) = h(y) · f (y) = h(y) · 0 = 0.



Proof (Cont.)

Now let N be an ideal such that My ⊂ N, My 6= N.

Then there existsg ∈ N such that g /∈ My . Thus, g(y) = c, where c 6= 0. Then if weconsider the constant function C , sending all elements of X to c , wenotice that

(g − C )(y) = g(y)− C (y) = c − c = 0.

Therefore, (g − C ) ∈ My ⊂ N. Since g is also in N, this implies that

g − (g − C ) = C ∈ N

Since c 6= 0, the constant function 1C sending every element of X to 1

c iswell-defined, and since N is an ideal, we have

1

C· C = 1 ∈ N,

implying that N = C (X ,F ). Since N was an arbitrary ideal properlycontaining My , this shows that My is maximal.


Proof (Cont.)

Now let N be an ideal such that My ⊂ N, My 6= N. Then there existsg ∈ N such that g /∈ My .

Thus, g(y) = c, where c 6= 0. Then if weconsider the constant function C , sending all elements of X to c , wenotice that

(g − C )(y) = g(y)− C (y) = c − c = 0.


g − (g − C ) = C ∈ N



1

C· C = 1 ∈ N,



Proof (Cont.)

Now let N be an ideal such that My ⊂ N, My 6= N. Then there existsg ∈ N such that g /∈ My . Thus, g(y) = c, where c 6= 0.

Then if weconsider the constant function C , sending all elements of X to c , wenotice that

(g − C )(y) = g(y)− C (y) = c − c = 0.


g − (g − C ) = C ∈ N



1

C· C = 1 ∈ N,



Proof (Cont.)

Now let N be an ideal such that My ⊂ N, My 6= N. Then there existsg ∈ N such that g /∈ My . Thus, g(y) = c, where c 6= 0. Then if weconsider the constant function C , sending all elements of X to c , wenotice that

(g − C )(y) = g(y)− C (y) = c − c = 0.


g − (g − C ) = C ∈ N



1

C· C = 1 ∈ N,



Proof (Cont.)


(g − C )(y) = g(y)− C (y) = c − c = 0.

Therefore, (g − C ) ∈ My ⊂ N.

Since g is also in N, this implies that

g − (g − C ) = C ∈ N



1

C· C = 1 ∈ N,



Proof (Cont.)


(g − C )(y) = g(y)− C (y) = c − c = 0.


g − (g − C ) = C ∈ N



1

C· C = 1 ∈ N,



Proof (Cont.)


(g − C )(y) = g(y)− C (y) = c − c = 0.


g − (g − C ) = C ∈ N



1

C· C = 1 ∈ N,



Proposition

Let R := C (X ,F ) and let

µ : X → Max (R)

be defined byµ(y) := My .

Regard Max (R) as a subspace of Spec (R) (endowed with Zariskitopology). Assume that {0} is a closed subset of F . Then µ is continuous.


Proof.

Recall that topology on Spec (R) is given by closed sets being exactly ofthe form V(E ) for some E ⊂ R, where

V(E ) := {P ∈ Spec (R) |E ⊂ P}.

Thus, as a subspace topology, the topology on Max (R) is given by closedsets being exactly of the form V(E ) for some E ⊂ R, where

V(E ) := {P ∈ Max (R) |E ⊂ P}.


Proof.

Recall that topology on Spec (R) is given by closed sets being exactly ofthe form V(E ) for some E ⊂ R, where

V(E ) := {P ∈ Spec (R) |E ⊂ P}.

Thus, as a subspace topology, the topology on Max (R) is given by closedsets being exactly of the form V(E ) for some E ⊂ R, where

V(E ) := {P ∈ Max (R) |E ⊂ P}.


Proof (Cont.)

Also, recall that this topology has as a basis sets of the form Df as franges over R, where

Df := Spec (R) \ V({f }) = {P ∈ Spec (R) |P does not contain f }.

Thus, as a subspace topology, Max (R) has basis elements of the form

Df := Max (R) \ V({f }) = {P ∈ Max (R) |P does not contain f }.

To show that µ is continuous it is enough to show that the inverse imageof a basis element is open.


Proof (Cont.)







Proof (Cont.)







Proof (Cont.)

Notice that for an arbitrary f ∈ R, we have:

µ−1(Df ) = {y ∈ X |µ(y) ∈ Df }= {y ∈ X | f /∈ My}= {y ∈ X | f (y) 6= 0}= f −1(F \ {0}).

(1)

Since by assumption {0} is a closed set in F , (F \ {0}) is open in F . Sincef is continuous, f −1(F \ {0}) is open in X . Thus, we can conclude thatthe inverse image of an arbitrary basis element is open, and therefore µ iscontinuous, as desired.


Proof (Cont.)



(1)

Since by assumption {0} is a closed set in F , (F \ {0}) is open in F . Sincef is continuous, f −1(F \ {0}) is open in X .

Thus, we can conclude thatthe inverse image of an arbitrary basis element is open, and therefore µ iscontinuous, as desired.


Proof (Cont.)



(1)

Since by assumption {0} is a closed set in F , (F \ {0}) is open in F . Sincef is continuous, f −1(F \ {0}) is open in X . Thus, we can conclude thatthe inverse image of an arbitrary basis element is open, and therefore µ iscontinuous, as desired.


Proposition

Assume that for each positive integer n, there is a polynomialPn ∈ F [X1, ...,Xn] such that

{(a1, ..., an) ∈ F n |Pn(a1, ..., an) = 0} = {(0, ..., 0)}.

Also, assume that {0} is a closed subset of F . Then the above map µ issurjective.

Proof.

Let M be a maximal ideal of R. Notice that if there exists y ∈ X suchthat f (y) = 0 for all f ∈ M, then M ⊂ My , which implies (by maximalityof M) that M = My . Thus, for contradiction assume that no such y exists,namely

∀ y ∈ X , ∃ fy ∈ M such that fy (y) 6= 0,

i.e. there is no point in the domain where all functions in M vanish.


Proposition


{(a1, ..., an) ∈ F n |Pn(a1, ..., an) = 0} = {(0, ..., 0)}.


Proof.

Let M be a maximal ideal of R. Notice that if there exists y ∈ X suchthat f (y) = 0 for all f ∈ M, then M ⊂ My , which implies (by maximalityof M) that M = My .

Thus, for contradiction assume that no such y exists,namely




Proposition


{(a1, ..., an) ∈ F n |Pn(a1, ..., an) = 0} = {(0, ..., 0)}.


Proof.

Let M be a maximal ideal of R. Notice that if there exists y ∈ X suchthat f (y) = 0 for all f ∈ M, then M ⊂ My , which implies (by maximalityof M) that M = My . Thus, for contradiction assume that no such y exists,namely




Proof (Cont.)

Since all functions in M are continuous and {0} is a closed subset of F ,for all y ∈ X , f −1

y (F \ {0}) is open and contains y . If we callBy := f −1

y (F \ {0}), then the above says that

∀ y ∈ X ,∃By such thatBy is open, y ∈ By , and fy (x) 6= 0∀ x ∈ By .

These open sets form an open cover of X , and since X is compact, we canfind a finite subcover, whose elements (after renaming) we can call{B1, ...,Bn}, together with the associated functions {f1, ..., fn}, where n issome positive integer.

By assumption, there exists a polynomialPn ∈ F [X1, ...,Xn] such that

{(a1, ..., an) ∈ F n |Pn(a1, ..., an) = 0} = {(0, ..., 0)},

i.e. a polynomial in n variables that only vanishes at the point {(0, ..., 0)}.


Proof (Cont.)

Since all functions in M are continuous and {0} is a closed subset of F ,for all y ∈ X , f −1

y (F \ {0}) is open and contains y . If we callBy := f −1

y (F \ {0}), then the above says that

∀ y ∈ X ,∃By such thatBy is open, y ∈ By , and fy (x) 6= 0∀ x ∈ By .

These open sets form an open cover of X , and since X is compact, we canfind a finite subcover, whose elements (after renaming) we can call{B1, ...,Bn}, together with the associated functions {f1, ..., fn}, where n issome positive integer. By assumption, there exists a polynomialPn ∈ F [X1, ...,Xn] such that

{(a1, ..., an) ∈ F n |Pn(a1, ..., an) = 0} = {(0, ..., 0)},

i.e. a polynomial in n variables that only vanishes at the point {(0, ..., 0)}.


Proof (Cont.)

Notice that since addition and multiplication functions as well as theidentity function are continuous maps in a topological field, polynomialsare continuous functions.

Thus, if g is the following composition offunctions:

X −→ F × ...× F −→ Fx −→ f1(x)× ...× fn(x) −→ Pn(f1(x), ..., fn(x)),

then g ∈ R. More specifically, since Pn(0, ..., 0) = 0, Pn does not have aconstant term, and therefore g ∈ M, for g is a combination (under ringaddition, ring multiplication, and multiplication of an element of an idealby an arbitrary element of a ring) of elements of M, namely f1, ..., fn.


Proof (Cont.)

Notice that since addition and multiplication functions as well as theidentity function are continuous maps in a topological field, polynomialsare continuous functions. Thus, if g is the following composition offunctions:


then g ∈ R.

More specifically, since Pn(0, ..., 0) = 0, Pn does not have aconstant term, and therefore g ∈ M, for g is a combination (under ringaddition, ring multiplication, and multiplication of an element of an idealby an arbitrary element of a ring) of elements of M, namely f1, ..., fn.


Proof (Cont.)

Notice that since addition and multiplication functions as well as theidentity function are continuous maps in a topological field, polynomialsare continuous functions. Thus, if g is the following composition offunctions:


then g ∈ R. More specifically, since Pn(0, ..., 0) = 0, Pn does not have aconstant term, and therefore g ∈ M, for g is a combination (under ringaddition, ring multiplication, and multiplication of an element of an idealby an arbitrary element of a ring) of elements of M, namely f1, ..., fn.


Proof (Cont.)

Also, notice that if g(x) = 0 for some x ∈ X , then by the property of Pn

we get that

g(x) = P(f1(x), ..., fn(x)) = 0 =⇒ f1(x) = 0, ..., fn(x) = 0,

which cannot be, since x ∈ Bi for some 1 ≤ i ≤ n, which in turn impliesthat fi (x) 6= 0. Therefore, g(x) 6= 0 for all x ∈ X .


Proof (Cont.)

However, if that is the case, then if 1g is the following composition of

functionsX −→ F \ {0} −→ Fx −→ g(x) −→ 1

g(x) ,

then 1g is a continuous function, being a composition of continuous

functions (where the second function is continuous by one of theproperties of a topological field).

Thus, since 1g ∈ R and g ∈ M, where M

is an ideal, we see that1

g· g = 1 ∈ M,

contradicting the maximality of M. This proves surjectivity of µ.


Proof (Cont.)

However, if that is the case, then if 1g is the following composition of

functionsX −→ F \ {0} −→ Fx −→ g(x) −→ 1

g(x) ,

then 1g is a continuous function, being a composition of continuous

functions (where the second function is continuous by one of theproperties of a topological field). Thus, since 1

g ∈ R and g ∈ M, where Mis an ideal, we see that

1

g· g = 1 ∈ M,

contradicting the maximality of M. This proves surjectivity of µ.


Proposition

Assume that given distinct x , y ∈ X , there is an f(x ,y) ∈ R such thatf(x ,y)(x) = 0 and f(x ,y)(y) 6= 0. Then the above map µ is injective.

Proof.

Let x , y ∈ X such that x 6= y . Then by assumption there is an f(x ,y) ∈ Rsuch that f(x ,y)(x) = 0 and f(x ,y)(y) 6= 0. Thus, f(x ,y) ∈ Mx , butf(x ,y) /∈ My . Therefore, Mx 6= My , which shows that µ is surjective.


Proposition

Assume that given distinct x , y ∈ X , there is an f(x ,y) ∈ R such thatf(x ,y)(x) = 0 and f(x ,y)(y) 6= 0. Then the above map µ is injective.

Proof.

Let x , y ∈ X such that x 6= y . Then by assumption there is an f(x ,y) ∈ Rsuch that f(x ,y)(x) = 0 and f(x ,y)(y) 6= 0. Thus, f(x ,y) ∈ Mx , butf(x ,y) /∈ My . Therefore, Mx 6= My , which shows that µ is surjective.


Theorem

Assume F = R. Then the above map µ is a homeomorphism.

Proof.

To prove that µ is a homeomorphism, it is enough to show that µ is aclosed map. Let U ⊂ X be a closed subset. We want to show that µ(U) isclosed in R. Now by Urysohn Lemma, for any y /∈ U there exists afunction fy : X → R such that f (u) = 0 for every u ∈ U and f (y) = 1.Knowing that, we will prove that

µ(U) = V({fy | y ∈ X \ U}

),

which will show that µ(U) is closed, proving that µ is a closed map, andthus a homeomorphism.


Theorem


Proof.

To prove that µ is a homeomorphism, it is enough to show that µ is aclosed map.

Let U ⊂ X be a closed subset. We want to show that µ(U) isclosed in R. Now by Urysohn Lemma, for any y /∈ U there exists afunction fy : X → R such that f (u) = 0 for every u ∈ U and f (y) = 1.Knowing that, we will prove that

µ(U) = V({fy | y ∈ X \ U}

),



Theorem


Proof.

To prove that µ is a homeomorphism, it is enough to show that µ is aclosed map. Let U ⊂ X be a closed subset. We want to show that µ(U) isclosed in R.

Now by Urysohn Lemma, for any y /∈ U there exists afunction fy : X → R such that f (u) = 0 for every u ∈ U and f (y) = 1.Knowing that, we will prove that

µ(U) = V({fy | y ∈ X \ U}

),



Theorem


Proof.

To prove that µ is a homeomorphism, it is enough to show that µ is aclosed map. Let U ⊂ X be a closed subset. We want to show that µ(U) isclosed in R. Now by Urysohn Lemma, for any y /∈ U there exists afunction fy : X → R such that f (u) = 0 for every u ∈ U and f (y) = 1.

Knowing that, we will prove that

µ(U) = V({fy | y ∈ X \ U}

),



Theorem


Proof.

To prove that µ is a homeomorphism, it is enough to show that µ is aclosed map. Let U ⊂ X be a closed subset. We want to show that µ(U) isclosed in R. Now by Urysohn Lemma, for any y /∈ U there exists afunction fy : X → R such that f (u) = 0 for every u ∈ U and f (y) = 1.Knowing that, we will prove that

µ(U) = V({fy | y ∈ X \ U}

),



Proof (Cont.)

Let Mx ∈ µ(U). Then Mx consists of all the functions that vanish on x .

Since all fy ’s vanish on entire U, they also vanish on x .Thus, Mx contains{fy | y ∈ X \ U}. Therefore, Mx ∈ V

({fy | y ∈ X \ U}

). Conversely, let

Mz ∈ V({fy | y ∈ X \ U}

)for some z ∈ X . We need to show that z ∈ U.

But notice that if z /∈ U, then by Urysohn Lemma there exists a functionfz : X → R such that fz(u) = 0 for every u ∈ U and f (z) = 1. Therefore,since fz ∈

({fy | y ∈ X \ U}

), fz ∈ Mz . But that cannot be, since Mz

consists of the functions that vanish on z . Thus, we have a contradiction,and consequently z ∈ U, which finishes the proof.


Proof (Cont.)

Let Mx ∈ µ(U). Then Mx consists of all the functions that vanish on x .Since all fy ’s vanish on entire U, they also vanish on x .Thus, Mx contains{fy | y ∈ X \ U}. Therefore, Mx ∈ V

({fy | y ∈ X \ U}

).

Conversely, letMz ∈ V

({fy | y ∈ X \ U}



({fy | y ∈ X \ U}




Proof (Cont.)


({fy | y ∈ X \ U}

). Conversely, let

Mz ∈ V({fy | y ∈ X \ U}

)for some z ∈ X .

We need to show that z ∈ U.But notice that if z /∈ U, then by Urysohn Lemma there exists a functionfz : X → R such that fz(u) = 0 for every u ∈ U and f (z) = 1. Therefore,since fz ∈

({fy | y ∈ X \ U}




Proof (Cont.)


({fy | y ∈ X \ U}

). Conversely, let

Mz ∈ V({fy | y ∈ X \ U}



({fy | y ∈ X \ U}




Proof (Cont.)


({fy | y ∈ X \ U}

). Conversely, let

Mz ∈ V({fy | y ∈ X \ U}


But notice that if z /∈ U, then by Urysohn Lemma there exists a functionfz : X → R such that fz(u) = 0 for every u ∈ U and f (z) = 1.

Therefore,since fz ∈

({fy | y ∈ X \ U}




Proof (Cont.)


({fy | y ∈ X \ U}

). Conversely, let

Mz ∈ V({fy | y ∈ X \ U}



({fy | y ∈ X \ U}

), fz ∈ Mz .

But that cannot be, since Mz



Proof (Cont.)


({fy | y ∈ X \ U}

). Conversely, let

Mz ∈ V({fy | y ∈ X \ U}



({fy | y ∈ X \ U}




Science

Properties of the set of continuous functions from a compact Hausdorff space to a topological field - Part I