properties of the set of continuous functions from a compact hausdorff space to a topological...

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

Example of an exotic 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 field part II – Example of an exotic topological fieldDecember 12, 2016 1 / 1

Recall the previously discussed definitions and propositions:

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}.

Proposition

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


Proposition

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

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.


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.

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.

Theorem

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


Remark

The assumption of compactness of our Hausdorff space is crucial.

To seethat, let X = (0, 1) and let F = R. Then for each n ∈ Z, n ≥ 1, considerthe following function:

fn(x) =

1 x ≤ n

n+1

linear nn+1 ≤ x ≤ 2n+1

2n+2

0 x ≥ 2n+12n+2

(1)


Remark

The assumption of compactness of our Hausdorff space is crucial. To seethat, let X = (0, 1) and let F = R.

Then for each n ∈ Z, n ≥ 1, considerthe following function:

fn(x) =

1 x ≤ n

n+1


2n+2

0 x ≥ 2n+12n+2

(1)


Remark

The assumption of compactness of our Hausdorff space is crucial. To seethat, let X = (0, 1) and let F = R. Then for each n ∈ Z, n ≥ 1, considerthe following function:

fn(x) =

1 x ≤ n

n+1


2n+2

0 x ≥ 2n+12n+2

(1)


Remark (Cont.)

Then consider the ideal generated by all such fn’s.

Call that ideal I . Thennotice that I 6= R, since if I = R, then 1 ∈ I , implying that

1 = g1fi1 + ...+ gmfim

for some g1, ..., gm ∈ R, fi1 , ..., fim ∈ I . But that cannot be, since if

N := max {i1, ..., im},

then for all x ≥ 2N+12N+2

fi1(x) = 0, ..., fim(x) = 0,

which in turn would imply that for any such x

1 = 1(x) = g1(x)fi1(x) + ...+ gm(x)fim(x) = 0,

a contradiction. Thus, I 6= R.


Remark (Cont.)

Then consider the ideal generated by all such fn’s. Call that ideal I .

Thennotice that I 6= R, since if I = R, then 1 ∈ I , implying that

1 = g1fi1 + ...+ gmfim


N := max {i1, ..., im},


fi1(x) = 0, ..., fim(x) = 0,


1 = 1(x) = g1(x)fi1(x) + ...+ gm(x)fim(x) = 0,



Remark (Cont.)

Then consider the ideal generated by all such fn’s. Call that ideal I . Thennotice that I 6= R, since if I = R, then 1 ∈ I , implying that

1 = g1fi1 + ...+ gmfim

for some g1, ..., gm ∈ R, fi1 , ..., fim ∈ I .

But that cannot be, since if

N := max {i1, ..., im},


fi1(x) = 0, ..., fim(x) = 0,


1 = 1(x) = g1(x)fi1(x) + ...+ gm(x)fim(x) = 0,



Remark (Cont.)


1 = g1fi1 + ...+ gmfim


N := max {i1, ..., im},


fi1(x) = 0, ..., fim(x) = 0,


1 = 1(x) = g1(x)fi1(x) + ...+ gm(x)fim(x) = 0,



Remark (Cont.)


1 = g1fi1 + ...+ gmfim


N := max {i1, ..., im},


fi1(x) = 0, ..., fim(x) = 0,


1 = 1(x) = g1(x)fi1(x) + ...+ gm(x)fim(x) = 0,

a contradiction.

Thus, I 6= R.


Remark (Cont.)


1 = g1fi1 + ...+ gmfim


N := max {i1, ..., im},


fi1(x) = 0, ..., fim(x) = 0,


1 = 1(x) = g1(x)fi1(x) + ...+ gm(x)fim(x) = 0,



Remark (Cont.)

Since R is a nonzero ring with identity, I ⊂ M, where M is a maximalideal.

However, notice that M is not of the form My for any y ∈ X . For ifthat was the case, then all functions in M would vanish at y . But for anyy , we can find n big enough so that fn(y) 6= 0. Since all fn’s are in M, wereach a contradiction. Thus, the previously mentioned map µ is notsurjective.


Remark (Cont.)

Since R is a nonzero ring with identity, I ⊂ M, where M is a maximalideal. However, notice that M is not of the form My for any y ∈ X .

For ifthat was the case, then all functions in M would vanish at y . But for anyy , we can find n big enough so that fn(y) 6= 0. Since all fn’s are in M, wereach a contradiction. Thus, the previously mentioned map µ is notsurjective.


Remark (Cont.)

Since R is a nonzero ring with identity, I ⊂ M, where M is a maximalideal. However, notice that M is not of the form My for any y ∈ X . For ifthat was the case, then all functions in M would vanish at y . But for anyy , we can find n big enough so that fn(y) 6= 0.

Since all fn’s are in M, wereach a contradiction. Thus, the previously mentioned map µ is notsurjective.


Remark (Cont.)

Since R is a nonzero ring with identity, I ⊂ M, where M is a maximalideal. However, notice that M is not of the form My for any y ∈ X . For ifthat was the case, then all functions in M would vanish at y . But for anyy , we can find n big enough so that fn(y) 6= 0. Since all fn’s are in M, wereach a contradiction.

Thus, the previously mentioned map µ is notsurjective.


Remark (Cont.)

Since R is a nonzero ring with identity, I ⊂ M, where M is a maximalideal. However, notice that M is not of the form My for any y ∈ X . For ifthat was the case, then all functions in M would vanish at y . But for anyy , we can find n big enough so that fn(y) 6= 0. Since all fn’s are in M, wereach a contradiction. Thus, the previously mentioned map µ is notsurjective.


Example

Let k be a field, let t be an indeterminate over k and let F := k((t)), i.e.the quotient field of the power series ring k[[t]].

It is a well-known factthat F coincides with the field of the formal Laurent series with coefficientfrom k defined by

k((t)) :=

∞∑

n≥Nant

n | an ∈ k , aN 6= 0, andN ∈ Z

∪ {0}Let ord denote the t-order, i.e.

ord

∞∑n≥N

antn

= N.


Example

Let k be a field, let t be an indeterminate over k and let F := k((t)), i.e.the quotient field of the power series ring k[[t]]. It is a well-known factthat F coincides with the field of the formal Laurent series with coefficientfrom k defined by

k((t)) :=

∞∑

n≥Nant

n | an ∈ k , aN 6= 0, andN ∈ Z

∪ {0}

Let ord denote the t-order, i.e.

ord

∞∑n≥N

antn

= N.


Example

Let k be a field, let t be an indeterminate over k and let F := k((t)), i.e.the quotient field of the power series ring k[[t]]. It is a well-known factthat F coincides with the field of the formal Laurent series with coefficientfrom k defined by

k((t)) :=

∞∑

n≥Nant

n | an ∈ k , aN 6= 0, andN ∈ Z

∪ {0}Let ord denote the t-order, i.e.

ord

∞∑n≥N

antn

= N.


Remark

Notice that the order of a Laurent series is the minimum index for whichthe coefficient of that series is non-zero. Therefore, under the assumptionthat the minimum of the empty set is infinity, the zero polynomial hasorder infinity.

Proposition

Given f , g ∈ F , define the function d : F × F → [0,∞) by

d(f , g) :=1

2ord(f−g).

Then d is a metric on F .


Proof.

First notice that the function f (x) = 12x is a non-negative function (it

attains zero at infinity). Therefore, d(f , g) ≥ 0 for all f , g ∈ F .

If f , g ∈ Fsuch that f = g , then ord(f − g) =∞, implying that d(f , g) = 0.Conversely, if d(f , g) = 0 for some f , g ∈ F , then ord(f − g) =∞,implying that f = g . To prove symmetry, notice that ord(f ) = ord(−f )for all f ∈ F , and thus

d(f , g) =1

2ord(f−g)=

1

2ord(−(f−g))=

1

2ord(g−f )= d(g , f )

Finally, we will present 2 proofs of the triangle inequality.


Proof.


attains zero at infinity). Therefore, d(f , g) ≥ 0 for all f , g ∈ F . If f , g ∈ Fsuch that f = g , then ord(f − g) =∞, implying that d(f , g) = 0.

Conversely, if d(f , g) = 0 for some f , g ∈ F , then ord(f − g) =∞,implying that f = g . To prove symmetry, notice that ord(f ) = ord(−f )for all f ∈ F , and thus

d(f , g) =1

2ord(f−g)=

1

2ord(−(f−g))=

1

2ord(g−f )= d(g , f )



Proof.


attains zero at infinity). Therefore, d(f , g) ≥ 0 for all f , g ∈ F . If f , g ∈ Fsuch that f = g , then ord(f − g) =∞, implying that d(f , g) = 0.Conversely, if d(f , g) = 0 for some f , g ∈ F , then ord(f − g) =∞,implying that f = g .

To prove symmetry, notice that ord(f ) = ord(−f )for all f ∈ F , and thus

d(f , g) =1

2ord(f−g)=

1

2ord(−(f−g))=

1

2ord(g−f )= d(g , f )



Proof.


attains zero at infinity). Therefore, d(f , g) ≥ 0 for all f , g ∈ F . If f , g ∈ Fsuch that f = g , then ord(f − g) =∞, implying that d(f , g) = 0.Conversely, if d(f , g) = 0 for some f , g ∈ F , then ord(f − g) =∞,implying that f = g . To prove symmetry, notice that ord(f ) = ord(−f )for all f ∈ F , and thus

d(f , g) =1

2ord(f−g)=

1

2ord(−(f−g))=

1

2ord(g−f )= d(g , f )



Proof (Cont.)

Version I. Recall that for any a, b ∈ F , we have the following property:

ord(a + b) ≥ min {ord(a), ord(b).} (2)

Thus, it follows that for any f , g , h ∈ F , we have

ord(f − g) = ord(f − h + h − g)

≥ min{ord(f − h), ord(h − g)

}.

(3)


Proof (Cont.)

Therefore,

d(f , g) =1

2ord(f−g)

≤ 1

2min{ord(f−h),ord(h−g)

}≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(4)


Proof (Cont.)

Version II. Let f , g , h ∈ F be arbitrary. To show the triangle inequality,we will consider 2 cases:

Case I: ord(f ) < ord(g). This implies that ord(f − g) = ord(f ). Also, insuch a case, ord(h) can be one of the following:

ord(f ) < ord(h). In this case, ord(f − h) = ord(f ), and thus

d(f , g) =1

2ord(f−g)

=1

2ord(f )

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(5)


Proof (Cont.)

Version II. Let f , g , h ∈ F be arbitrary. To show the triangle inequality,we will consider 2 cases:Case I: ord(f ) < ord(g).

This implies that ord(f − g) = ord(f ). Also, insuch a case, ord(h) can be one of the following:


d(f , g) =1

2ord(f−g)

=1

2ord(f )

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(5)


Proof (Cont.)

Version II. Let f , g , h ∈ F be arbitrary. To show the triangle inequality,we will consider 2 cases:Case I: ord(f ) < ord(g). This implies that ord(f − g) = ord(f ).

Also, insuch a case, ord(h) can be one of the following:


d(f , g) =1

2ord(f−g)

=1

2ord(f )

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(5)


Proof (Cont.)

Version II. Let f , g , h ∈ F be arbitrary. To show the triangle inequality,we will consider 2 cases:Case I: ord(f ) < ord(g). This implies that ord(f − g) = ord(f ). Also, insuch a case, ord(h) can be one of the following:

ord(f ) < ord(h).

In this case, ord(f − h) = ord(f ), and thus

d(f , g) =1

2ord(f−g)

=1

2ord(f )

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(5)


Proof (Cont.)


ord(f ) < ord(h). In this case, ord(f − h) = ord(f ),

and thus

d(f , g) =1

2ord(f−g)

=1

2ord(f )

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(5)


Proof (Cont.)



d(f , g) =1

2ord(f−g)

=1

2ord(f )

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(5)


Proof (Cont.)

ord(f ) > ord(h).

In this case, ord(f − h) = ord(h), and thus

d(f , g) =1

2ord(f−g)

=1

2ord(f )

<1

2ord(h)

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(6)


Proof (Cont.)

ord(f ) > ord(h). In this case, ord(f − h) = ord(h),

and thus

d(f , g) =1

2ord(f−g)

=1

2ord(f )

<1

2ord(h)

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(6)


Proof (Cont.)

ord(f ) > ord(h). In this case, ord(f − h) = ord(h), and thus

d(f , g) =1

2ord(f−g)

=1

2ord(f )

<1

2ord(h)

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(6)


Proof (Cont.)

ord(f ) = ord(h).

In this case, ord(f ) = ord(f − g) = ord(h− g), andthus

d(f , g) =1

2ord(f−g)

=1

2ord(h−g)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(7)


Proof (Cont.)

ord(f ) = ord(h). In this case, ord(f ) = ord(f − g) = ord(h− g),

andthus

d(f , g) =1

2ord(f−g)

=1

2ord(h−g)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(7)


Proof (Cont.)

ord(f ) = ord(h). In this case, ord(f ) = ord(f − g) = ord(h− g), andthus

d(f , g) =1

2ord(f−g)

=1

2ord(h−g)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(7)


Proof (Cont.)

Case II: ord(f ) = ord(g).

This implies that ord(f − g) ≥ ord(f ). Also, insuch a case, ord(h) can be one of the following:


d(f , g) =1

2ord(f−g)

≤ 1

2ord(f )

=1

2ord(f−h)

<1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(8)


Proof (Cont.)

Case II: ord(f ) = ord(g). This implies that ord(f − g) ≥ ord(f ).

Also, insuch a case, ord(h) can be one of the following:


d(f , g) =1

2ord(f−g)

≤ 1

2ord(f )

=1

2ord(f−h)

<1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(8)


Proof (Cont.)

Case II: ord(f ) = ord(g). This implies that ord(f − g) ≥ ord(f ). Also, insuch a case, ord(h) can be one of the following:

ord(f ) < ord(h).

In this case, ord(f − h) = ord(f ), and thus

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f )

=1

2ord(f−h)

<1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(8)


Proof (Cont.)


ord(f ) < ord(h). In this case, ord(f − h) = ord(f ),

and thus

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f )

=1

2ord(f−h)

<1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(8)


Proof (Cont.)



d(f , g) =1

2ord(f−g)

≤ 1

2ord(f )

=1

2ord(f−h)

<1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(8)


Proof (Cont.)

ord(f ) > ord(h).

In this case, ord(f − h) = ord(h), and thus

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f )

<1

2ord(h)

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(9)


Proof (Cont.)

ord(f ) > ord(h). In this case, ord(f − h) = ord(h),

and thus

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f )

<1

2ord(h)

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(9)


Proof (Cont.)

ord(f ) > ord(h). In this case, ord(f − h) = ord(h), and thus

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f )

<1

2ord(h)

=1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(9)


Proof (Cont.)

ord(f ) = ord(h).

In this case, let n := ord(f ) = ord(h) = ord(g) andlet k := ord(f − g). Then k ≥ n. That means that f and g share thefirst k − n coefficients. Now, if ord(f − h) ≤ ord(f − g) = k, then

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(10)


Proof (Cont.)

ord(f ) = ord(h). In this case, let n := ord(f ) = ord(h) = ord(g) andlet k := ord(f − g).

Then k ≥ n. That means that f and g share thefirst k − n coefficients. Now, if ord(f − h) ≤ ord(f − g) = k, then

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(10)


Proof (Cont.)

ord(f ) = ord(h). In this case, let n := ord(f ) = ord(h) = ord(g) andlet k := ord(f − g). Then k ≥ n.

That means that f and g share thefirst k − n coefficients. Now, if ord(f − h) ≤ ord(f − g) = k, then

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(10)


Proof (Cont.)

ord(f ) = ord(h). In this case, let n := ord(f ) = ord(h) = ord(g) andlet k := ord(f − g). Then k ≥ n. That means that f and g share thefirst k − n coefficients.

Now, if ord(f − h) ≤ ord(f − g) = k, then

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(10)


Proof (Cont.)

ord(f ) = ord(h). In this case, let n := ord(f ) = ord(h) = ord(g) andlet k := ord(f − g). Then k ≥ n. That means that f and g share thefirst k − n coefficients. Now, if ord(f − h) ≤ ord(f − g) = k, then

d(f , g) =1

2ord(f−g)

≤ 1

2ord(f−h)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

(10)


Proof (Cont.)

If k ′ := ord(f − h) > ord(f − g) = k , then f and g share the first k ′ − ncoefficients.

In other words, f shares more first coefficients with h thanwith g . Since the first k ′ − n coefficents of f and h are the same, h sharesexactly as many first coefficients with g as f does. Thus,ord(f − g) = ord(h − g) and we can conclude that

d(f , g) =1

2ord(f−g)

=1

2ord(h−g)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

Remark

The case when ord(f ) > ord(g) follows from case I by symmetry.


Proof (Cont.)

If k ′ := ord(f − h) > ord(f − g) = k , then f and g share the first k ′ − ncoefficients. In other words, f shares more first coefficients with h thanwith g .

Since the first k ′ − n coefficents of f and h are the same, h sharesexactly as many first coefficients with g as f does. Thus,ord(f − g) = ord(h − g) and we can conclude that

d(f , g) =1

2ord(f−g)

=1

2ord(h−g)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

Remark



Proof (Cont.)

If k ′ := ord(f − h) > ord(f − g) = k , then f and g share the first k ′ − ncoefficients. In other words, f shares more first coefficients with h thanwith g . Since the first k ′ − n coefficents of f and h are the same, h sharesexactly as many first coefficients with g as f does.

Thus,ord(f − g) = ord(h − g) and we can conclude that

d(f , g) =1

2ord(f−g)

=1

2ord(h−g)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

Remark



Proof (Cont.)

If k ′ := ord(f − h) > ord(f − g) = k , then f and g share the first k ′ − ncoefficients. In other words, f shares more first coefficients with h thanwith g . Since the first k ′ − n coefficents of f and h are the same, h sharesexactly as many first coefficients with g as f does. Thus,ord(f − g) = ord(h − g)

and we can conclude that

d(f , g) =1

2ord(f−g)

=1

2ord(h−g)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

Remark



Proof (Cont.)

If k ′ := ord(f − h) > ord(f − g) = k , then f and g share the first k ′ − ncoefficients. In other words, f shares more first coefficients with h thanwith g . Since the first k ′ − n coefficents of f and h are the same, h sharesexactly as many first coefficients with g as f does. Thus,ord(f − g) = ord(h − g) and we can conclude that

d(f , g) =1

2ord(f−g)

=1

2ord(h−g)

≤ 1

2ord(f−h)+

1

2ord(h−g)

= d(f , h) + d(h, g).

Remark



Proposition

Under the metric topology resulting from the above metric, F is atopological field.

Proof.

To show that F is a topological field, we need to show that addition,multiplication, and reciprocal maps are continuous.


Proof (Cont.)

Case I: To show that addition is continuous, let (fn, gn)n→∞−−−→ (f , g) in

the product topology induced by the above metric.

Recall that thishappens if and only if

fnn→∞−−−→ f and gn

n→∞−−−→ g .

Notice that this is equivalent to saying that

d(fn, f )n→∞−−−→ 0 and d(gn, g)

n→∞−−−→ 0.

But from the definition of our metric, this is equivalent to saying that

ord(fn − f )n→∞−−−→∞ and ord(gn − g)

n→∞−−−→∞,

i.e. fn (respectively gn) has more and more ”first” coefficients exactlysame as f (respectively g) as n goes to infinity.


Proof (Cont.)

Case I: To show that addition is continuous, let (fn, gn)n→∞−−−→ (f , g) in

the product topology induced by the above metric. Recall that thishappens if and only if

fnn→∞−−−→ f and gn

n→∞−−−→ g .

Notice that this is equivalent to saying that

d(fn, f )n→∞−−−→ 0 and d(gn, g)

n→∞−−−→ 0.

But from the definition of our metric, this is equivalent to saying that

ord(fn − f )n→∞−−−→∞ and ord(gn − g)

n→∞−−−→∞,

i.e. fn (respectively gn) has more and more ”first” coefficients exactlysame as f (respectively g) as n goes to infinity.


Proof (Cont.)

Thus, from the above discussion, to show that

fn + gnn→∞−−−→ f + g ,

it is enough to show that

ord(fn + gn − (f + g))n→∞−−−→∞.


Proof (Cont.)

To do so, recall that for any a, b ∈ F , we have the following property:

ord(a + b) ≥ min {ord(a), ord(b)} (11)

Therefore, we see that

ord(fn + gn − (f + g)) = ord(fn − f + g − gn)

≥ min {ord(fn − f ), ord(g − gn)}= min {ord(fn − f ), ord(gn − g)},

(12)

and since both orders on the right go to infinity as n goes to infinity, sodoes the minimum, and thus as n goes to infinity, the order offn + gn − (f + g) goes to infinity as well.


Proof (Cont.)

Case II: To show that multiplication is continuous, let fn and gn be asabove.

By the above discussion, it is enough to show that

ord(fngn − fg)n→∞−−−→∞

To do so, notice that if gnn→∞−−−→ g , then ord(gn − g)

n→∞−−−→∞, which inturn implies that if g 6= 0, then for a sufficiently large n,

ord(gn) = ord(g),

and if g = 0, thenord(gn)

n→∞−−−→∞.

Also, recall that since F is a field (and thus an integral domain), it has nozero divisors, and therefore for any a, b ∈ F ,

ord(ab) = ord(a) + ord(b) (13)


Proof (Cont.)

Case II: To show that multiplication is continuous, let fn and gn be asabove. By the above discussion, it is enough to show that




ord(gn) = ord(g),


n→∞−−−→∞.




Proof (Cont.)




n→∞−−−→∞,

which inturn implies that if g 6= 0, then for a sufficiently large n,

ord(gn) = ord(g),


n→∞−−−→∞.




Proof (Cont.)





ord(gn) = ord(g),


n→∞−−−→∞.




Proof (Cont.)

These two facts help us notice that

ord(fngn − fg) = ord(fngn − fgn + fgn − fg)

= ord((fn − f )gn + f (gn − g))

≥ min {ord((fn − f )gn), ord(f (gn − g))},≥ min {ord(fn − f ) + ord(gn), ord(f ) + ord(gn − g)}.

(14)

But as n goes to infinity, ord(gn) becomes fixed or goes to infinity, ord(f )is fixed, and both ord(fn − f ) and ord(gn − g) go to infinity. Therefore,this implies that

ord(fngn − fg)n→∞−−−→∞.


Proof (Cont.)

Case III: To show that the reciprocal map is continuous, let fnn→∞−−−→ f ,

where f 6= 0 and fn 6= 0 for all n.

By the above discussion, it is enough toshow that

ord

(1

fn− 1

f

)n→∞−−−→∞

Recall that since F is a field (and thus an integral domain), it has no zerodivisors, and therefore for any a, b ∈ F such that b 6= 0, we have

ord(ab

)= ord(a)− ord(b) (15)


Proof (Cont.)

Case III: To show that the reciprocal map is continuous, let fnn→∞−−−→ f ,

where f 6= 0 and fn 6= 0 for all n. By the above discussion, it is enough toshow that

ord

(1

fn− 1

f

)n→∞−−−→∞

Recall that since F is a field (and thus an integral domain), it has no zerodivisors, and therefore for any a, b ∈ F such that b 6= 0, we have

ord(ab

)= ord(a)− ord(b) (15)


Proof (Cont.)

Therefore, by using this fact and facts established before, we see that

ord

(1

fn− 1

f

)= ord

(f − fnfnf

)= ord(fn − f )− ord(fnf )

= ord(fn − f )− (ord(fn) + ord(f ))

(16)

Since f 6= 0, order of fn becomes fixed for big enough n and ord(fn − f )goes to infinity, the left side also approaches infinity an n increases. Thisfinishes the proof that F is a topological field.


Proposition

Let P2 := X 22 − tX 2

1 and inductively define

Pn := X 2n − tP2

n−1 for all integers n ≥ 3

Then for (a1, ..., an) ∈ F n, we have

Pn(a1, ..., an) = 0 iff (a1, ..., an) = (0, ..., 0).


Proof.

First, let (a1, ..., an) = (0, ..., 0). Then we proceed by induction:

Base case: If k = 2, then

P2(0, 0) = 02 − t · 02 = 0.

Inductive step: Assume that the statement holds for k − 1 < n. Then

Pk(0, ..., 0) = 02 − t(Pk−1(0, ..., 0)

)2= 0.


Proof.

First, let (a1, ..., an) = (0, ..., 0). Then we proceed by induction:Base case: If k = 2, then

P2(0, 0) = 02 − t · 02 = 0.

Inductive step: Assume that the statement holds for k − 1 < n. Then

Pk(0, ..., 0) = 02 − t(Pk−1(0, ..., 0)

)2= 0.


Proof (Cont.)

Conversely, let (a1, ..., an) be such that Pn(a1, ..., an) = 0.

Then

Pn(a1, ..., an) = a2n − t

(Pn−1(a1, ..., an−1)

)2= 0,

which is equivalent to

a2n = t

(Pn−1(a1, ..., an−1)

)2.

However, if the order of an is finite, we see that the t-order on the left sideis even, whereas the t-order on the right side is odd. Thus, order has to beinfinite, implying that an = 0 and Pn−1(a1, ..., an−1) = 0.


Proof (Cont.)

Conversely, let (a1, ..., an) be such that Pn(a1, ..., an) = 0. Then

Pn(a1, ..., an) = a2n − t

(Pn−1(a1, ..., an−1)

)2= 0,


a2n = t

(Pn−1(a1, ..., an−1)

)2.



Proof (Cont.)


Pn(a1, ..., an) = a2n − t

(Pn−1(a1, ..., an−1)

)2= 0,


a2n = t

(Pn−1(a1, ..., an−1)

)2.

However, if the order of an is finite, we see that the t-order on the left sideis even, whereas the t-order on the right side is odd.

Thus, order has to beinfinite, implying that an = 0 and Pn−1(a1, ..., an−1) = 0.


Proof (Cont.)


Pn(a1, ..., an) = a2n − t

(Pn−1(a1, ..., an−1)

)2= 0,


a2n = t

(Pn−1(a1, ..., an−1)

)2.



Proof (Cont.)

Continuing in this manner, we get that

(an, ..., a3) = 0 and P2(a1, a2) = a22 − ta2

1 = 0

But by applying the same logic, we see that (a1, a2) = (0, 0), implying that

(a1, ..., an) = (0, ..., 0),

as desired.


properties of the set of continuous functions from a compact hausdorff space to a topological...

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