Tancuj, tancuj, konvergujJako v konvergenci i v tanci nezáleží na pár prvních

krocích, ale jak tancujete ke konci a kam dotancujete.Photo by tyle_r on Flickr

Které vlastnosti jdou vyjádřit konvergencí

•Kompaktnost•Úplnost•Totální omezenost•Omezenost•Otevřenost•Spojitost•Stejnoměrná spojitost• Souvislost• Separabilnost

Které vlastnosti jdou vyjádřit konvergencí

1. Uzavřenost na podposloupnosti2. Jednoznačnost limity3. Posloupnost nekonvergující k x obsahuje

podposloupnost, která je “daleko od x”.4. Konvergence konstantní posloupnosti

Topologie

•Kompaktnost

•Úplnost

•Totální omezenost

•Omezenost

•Otevřenost

•Spojitost

•Stejnoměrná spojitost

•Souvislost

•Separabilnost

1. Podposloupnosti blízkých jsou blízké2. Různé konstantní nejsou blízké3. Neblízké posloupnosti obsahují

podposloupnosti, které jsou“daleko od sebe”.

Cauchyovská posloupnost

Kam jsme dotancovali

Dirty Dancing (1987)

•Kompaktnost

•Úplnost

•Totální omezenost

•Omezenost

•Otevřenost

•Spojitost

•Stejnoměrná spojitost

•Souvislost

•Separabilnost

3. Properties defined byconvergence

In this chapter we will use the following convention: when we say a uniformlysequential space (X,⇠) has a property defined for a sequential space we meanthat ⌅⇠ has this property.

3.1 Continuous and uniformly continuous map-

pings

Definition 3.1. Let ⌅X

and ⌅Y

be two sequential structures on sets X and Y .We say f : X ! Y is a continuous mapping ⌅

X

to ⌅Y

if ({xn

}, x) 2 ⌅X

impliesthat ({f(x

n

)}, f(x)) 2 ⌅Y

.

The following theorem is often covered in basic course of mathematical ana-lysis. We include it for completeness.

Theorem 3.2. Let ⌅X

and ⌅Y

be two sequential structures generated by metricspaces (X, %) and (Y, �). Let f : X ! Y be a mapping. Then the following areequivalent:

(i) f : (X, %) ! (Y, �) is continuous;

(ii) f : ⌅X

! ⌅Y

is continuous.

Proof. Let x 2 X be a fixed point.Let condition (i) hold and x

n

! x (that is, ({xn

}, x) 2 ⌅X

). For " > 0 wefind � > 0 from Definition 1.6. Then there exists n

"

such that for 8n � n

"

wehave %(x

n

, x) < � and then �(f(xn

), f(x)) < ". Hence f(xn

) ! f(x) in Y , thatis, ({f(x

n

)}, f(x)) 2 ⌅Y

.Let now condition (ii) hold and for contradiction suppose that (i) does not.

Then9" > 0 8� > 0 8y 2 X : %(x, y) < � ) �(f(x), f(y)) � ".

We fix this " and find a sequence {xn

} such that %(x1, x) < 1 and %(xn+1, x) <

min( 1n

, %(xn

, x)) for n 2 N. Then x

n

! x hence from (ii) f(xn

) ! f(x). This isnot possible since �(f(x

n

), f(x)) � " for n 2 N.

Definition 3.3. Let (X,⇠X

) and (Y,⇠Y

) be two uniformly sequential spaces.We say f : X ! Y is a uniformly continuous mapping ⇠

X

to ⇠Y

if {xn

} ⇠X

{yn

}implies that {f(x

n

)} ⇠Y

{f(yn

)}.

Theorem 3.4. Let ⇠X

and ⇠Y

be two uniformly sequential structures generatedby metric spaces (X, %) and (Y, �). Let f : X ! Y be a mapping. Then thefollowing are equivalent:

(i) f : (X, %) ! (Y, �) is uniformly continuous;

(ii) f : (X,⇠X

) ! (Y,⇠Y

) is uniformly continuous.

10

3. Properties defined byconvergence

In this chapter we will use the following convention: when we say a uniformlysequential space (X,⇠) has a property defined for a sequential space we meanthat ⌅⇠ has this property.

3.1 Continuous and uniformly continuous map-

pings

Definition 3.1. Let ⌅X

and ⌅Y

be two sequential structures on sets X and Y .We say f : X ! Y is a continuous mapping ⌅

X

to ⌅Y

if ({xn

}, x) 2 ⌅X

impliesthat ({f(x

n

)}, f(x)) 2 ⌅Y

.

The following theorem is often covered in basic course of mathematical ana-lysis. We include it for completeness.

Theorem 3.2. Let ⌅X

and ⌅Y

be two sequential structures generated by metricspaces (X, %) and (Y, �). Let f : X ! Y be a mapping. Then the following areequivalent:

(i) f : (X, %) ! (Y, �) is continuous;

(ii) f : ⌅X

! ⌅Y

is continuous.

Proof. Let x 2 X be a fixed point.Let condition (i) hold and x

n

! x (that is, ({xn

}, x) 2 ⌅X

). For " > 0 wefind � > 0 from Definition 1.6. Then there exists n

"

such that for 8n � n

"

wehave %(x

n

, x) < � and then �(f(xn

), f(x)) < ". Hence f(xn

) ! f(x) in Y , thatis, ({f(x

n

)}, f(x)) 2 ⌅Y

.Let now condition (ii) hold and for contradiction suppose that (i) does not.

Then9" > 0 8� > 0 8y 2 X : %(x, y) < � ) �(f(x), f(y)) � ".

We fix this " and find a sequence {xn

} such that %(x1, x) < 1 and %(xn+1, x) <

min( 1n

, %(xn

, x)) for n 2 N. Then x

n

! x hence from (ii) f(xn

) ! f(x). This isnot possible since �(f(x

n

), f(x)) � " for n 2 N.

Definition 3.3. Let (X,⇠X

) and (Y,⇠Y

) be two uniformly sequential spaces.We say f : X ! Y is a uniformly continuous mapping ⇠

X

to ⇠Y

if {xn

} ⇠X

{yn

}implies that {f(x

n

)} ⇠Y

{f(yn

)}.

Theorem 3.4. Let ⇠X

and ⇠Y

be two uniformly sequential structures generatedby metric spaces (X, %) and (Y, �). Let f : X ! Y be a mapping. Then thefollowing are equivalent:

(i) f : (X, %) ! (Y, �) is uniformly continuous;

(ii) f : (X,⇠X

) ! (Y,⇠Y

) is uniformly continuous.

10

3. Properties defined byconvergence

In this chapter we will use the following convention: when we say a uniformlysequential space (X,⇠) has a property defined for a sequential space we meanthat ⌅⇠ has this property.

3.1 Continuous and uniformly continuous map-

pings

Definition 3.1. Let ⌅X

and ⌅Y

be two sequential structures on sets X and Y .We say f : X ! Y is a continuous mapping ⌅

X

to ⌅Y

if ({xn

}, x) 2 ⌅X

impliesthat ({f(x

n

)}, f(x)) 2 ⌅Y

.

The following theorem is often covered in basic course of mathematical ana-lysis. We include it for completeness.

Theorem 3.2. Let ⌅X

and ⌅Y

be two sequential structures generated by metricspaces (X, %) and (Y, �). Let f : X ! Y be a mapping. Then the following areequivalent:

(i) f : (X, %) ! (Y, �) is continuous;

(ii) f : ⌅X

! ⌅Y

is continuous.

Proof. Let x 2 X be a fixed point.Let condition (i) hold and x

n

! x (that is, ({xn

}, x) 2 ⌅X

). For " > 0 wefind � > 0 from Definition 1.6. Then there exists n

"

such that for 8n � n

"

wehave %(x

n

, x) < � and then �(f(xn

), f(x)) < ". Hence f(xn

) ! f(x) in Y , thatis, ({f(x

n

)}, f(x)) 2 ⌅Y

.Let now condition (ii) hold and for contradiction suppose that (i) does not.

Then9" > 0 8� > 0 8y 2 X : %(x, y) < � ) �(f(x), f(y)) � ".

We fix this " and find a sequence {xn

} such that %(x1, x) < 1 and %(xn+1, x) <

min( 1n

, %(xn

, x)) for n 2 N. Then x

n

! x hence from (ii) f(xn

) ! f(x). This isnot possible since �(f(x

n

), f(x)) � " for n 2 N.

Definition 3.3. Let (X,⇠X

) and (Y,⇠Y

) be two uniformly sequential spaces.We say f : X ! Y is a uniformly continuous mapping ⇠

X

to ⇠Y

if {xn

} ⇠X

{yn

}implies that {f(x

n

)} ⇠Y

{f(yn

)}.

Theorem 3.4. Let ⇠X

and ⇠Y

be two uniformly sequential structures generatedby metric spaces (X, %) and (Y, �). Let f : X ! Y be a mapping. Then thefollowing are equivalent:

(i) f : (X, %) ! (Y, �) is uniformly continuous;

(ii) f : (X,⇠X

) ! (Y,⇠Y

) is uniformly continuous.

10

3. Properties defined byconvergence

In this chapter we will use the following convention: when we say a uniformlysequential space (X,⇠) has a property defined for a sequential space we meanthat ⌅⇠ has this property.

3.1 Continuous and uniformly continuous map-

pings

Definition 3.1. Let ⌅X

and ⌅Y

be two sequential structures on sets X and Y .We say f : X ! Y is a continuous mapping ⌅

X

to ⌅Y

if ({xn

}, x) 2 ⌅X

impliesthat ({f(x

n

)}, f(x)) 2 ⌅Y

.

The following theorem is often covered in basic course of mathematical ana-lysis. We include it for completeness.

Theorem 3.2. Let ⌅X

and ⌅Y

be two sequential structures generated by metricspaces (X, %) and (Y, �). Let f : X ! Y be a mapping. Then the following areequivalent:

(i) f : (X, %) ! (Y, �) is continuous;

(ii) f : ⌅X

! ⌅Y

is continuous.

Proof. Let x 2 X be a fixed point.Let condition (i) hold and x

n

! x (that is, ({xn

}, x) 2 ⌅X

). For " > 0 wefind � > 0 from Definition 1.6. Then there exists n

"

such that for 8n � n

"

wehave %(x

n

, x) < � and then �(f(xn

), f(x)) < ". Hence f(xn

) ! f(x) in Y , thatis, ({f(x

n

)}, f(x)) 2 ⌅Y

.Let now condition (ii) hold and for contradiction suppose that (i) does not.

Then9" > 0 8� > 0 8y 2 X : %(x, y) < � ) �(f(x), f(y)) � ".

We fix this " and find a sequence {xn

} such that %(x1, x) < 1 and %(xn+1, x) <

min( 1n

, %(xn

, x)) for n 2 N. Then x

n

! x hence from (ii) f(xn

) ! f(x). This isnot possible since �(f(x

n

), f(x)) � " for n 2 N.

Definition 3.3. Let (X,⇠X

) and (Y,⇠Y

) be two uniformly sequential spaces.We say f : X ! Y is a uniformly continuous mapping ⇠

X

to ⇠Y

if {xn

} ⇠X

{yn

}implies that {f(x

n

)} ⇠Y

{f(yn

)}.

Theorem 3.4. Let ⇠X

and ⇠Y

be two uniformly sequential structures generatedby metric spaces (X, %) and (Y, �). Let f : X ! Y be a mapping. Then thefollowing are equivalent:

(i) f : (X, %) ! (Y, �) is uniformly continuous;

(ii) f : (X,⇠X

) ! (Y,⇠Y

) is uniformly continuous.

10

∃

Theorem 3.16. Let (X,⇠X

), (Y,⇠Y

) be two uniformly sequential spaces, f :X ! Y be a continuous mapping and (X,⇠

X

) be compact. Then f is uniformlycontinuous.

Proof. Let {xn

} ⇠X

{yn

} and let us for contradiction suppose that {f(xn

)} 6⇠Y

{f(yn

)}. Then from (U3) we find subsequences {f(xnk)} and {f(y

nk)} such

that for none their subsequences we have {f(xnki

)} ⇠Y

{f(ynki

)}. Because X

is compact we can find subsequences {xnki

} of {xnk}, {y

nki} of {y

nk} and points

x, y 2 X such that ({xnki

}, x) 2 ⌅⇠X and ({ynki

}, y) 2 ⌅⇠X . Since {xn

} ⇠X

{yn

}we have {x

nk} ⇠

X

{ynk} and therefore x = y. Mapping f is continuous so

({f(xnki

)}, f(x)) 2 ⌅⇠Y and ({f(ynki

)}, f(x)) 2 ⌅⇠Y . That is, {f(xnki

)} ⇠Y

{f(x)} and {f(ynki

)} ⇠Y

{f(x)} which gives us {f(xnki

)} ⇠Y

{f(ynki

)} which isthe contradiction.

The following example proves that the implication in the previous theoremcan not be reversed.

Example 3.17. We say two infinite sets are almost disjoint if their intersectionis finite. Let us assume an infinite maximal family of almost disjoint subsets ofN and denote it as MAD(N).

For a reader’s image of some infinite family of almost disjoint subsets of N wewill show a construction of one. Let f : N ! Q be a bijection of natural numbersonto rational numbers. For every irrational x 2 R \Q we will choose one rationalsequence {r

n

} such that r

n

! x. Then Sx

:= f

�1[{rn

}] ⇢ N is a preimage ofthat sequence. Then {S

x

: x 2 R \ Q} is an infinite family of almost disjointsubsets of N because if two sets had an infinite intersection the correspondingrational sequences would have an infinite intersection too and since they are bothconvergent it follows they would converge to the same point.

Please note that using this construction we do not obtain a MAD(N) as forexample the preimage of the set of all even numbers is almost disjoint with allconstructed sets.

We denote MAD(N) = {S↵

: ↵ 2 A} for some set A. Let X := N [MAD(N)with the smallest sequential structure such that

S↵

! {S↵

}, {S↵

} 2 MAD(N).

This condition can be rephrased: members of S↵

as points in N converge to S↵

as point in MAD(N). Then every continuous mapping to R is bounded but thespace is evidently not compact.

To prove this let f : X ! R be a continuous unbounded mapping. That is,there exists a sequence {x

n

} for which f(xn

) ! 1. Now we examine two cases.First, infinitely many members of {x

n

} lie in N, so we can find a subsequence{x

nk} ⇢ N. Then there exists S

↵

2 MAD(N) such that {xnk} ! S

↵

. Then wehave {f(x

nk)} ! f(S

↵

) and {f(xnk)} ! 1 which is a contradiction.

Second, only finite number of members of {xn

} lie in N. Without loss ofgenerality we suppose that {x

n

} = {S↵n} ⇢ MAD(N). For every n 2 N we can

find y

n

2 N such that |f(yn

) � f(S↵n)| 1. The first case yields {f(y

n

)} isbounded and so {f(S

↵n)} is bounded.

14

Theorem 3.16. Let (X,⇠X

), (Y,⇠Y

) be two uniformly sequential spaces, f :X ! Y be a continuous mapping and (X,⇠

X

) be compact. Then f is uniformlycontinuous.

Proof. Let {xn

} ⇠X

{yn

} and let us for contradiction suppose that {f(xn

)} 6⇠Y

{f(yn

)}. Then from (U3) we find subsequences {f(xnk)} and {f(y

nk)} such

that for none their subsequences we have {f(xnki

)} ⇠Y

{f(ynki

)}. Because X

is compact we can find subsequences {xnki

} of {xnk}, {y

nki} of {y

nk} and points

x, y 2 X such that ({xnki

}, x) 2 ⌅⇠X and ({ynki

}, y) 2 ⌅⇠X . Since {xn

} ⇠X

{yn

}we have {x

nk} ⇠

X

{ynk} and therefore x = y. Mapping f is continuous so

({f(xnki

)}, f(x)) 2 ⌅⇠Y and ({f(ynki

)}, f(x)) 2 ⌅⇠Y . That is, {f(xnki

)} ⇠Y

{f(x)} and {f(ynki

)} ⇠Y

{f(x)} which gives us {f(xnki

)} ⇠Y

{f(ynki

)} which isthe contradiction.

The following example proves that the implication in the previous theoremcan not be reversed.

Example 3.17. We say two infinite sets are almost disjoint if their intersectionis finite. Let us assume an infinite maximal family of almost disjoint subsets ofN and denote it as MAD(N).

For a reader’s image of some infinite family of almost disjoint subsets of N wewill show a construction of one. Let f : N ! Q be a bijection of natural numbersonto rational numbers. For every irrational x 2 R \Q we will choose one rationalsequence {r

n

} such that r

n

! x. Then Sx

:= f

�1[{rn

}] ⇢ N is a preimage ofthat sequence. Then {S

x

: x 2 R \ Q} is an infinite family of almost disjointsubsets of N because if two sets had an infinite intersection the correspondingrational sequences would have an infinite intersection too and since they are bothconvergent it follows they would converge to the same point.

Please note that using this construction we do not obtain a MAD(N) as forexample the preimage of the set of all even numbers is almost disjoint with allconstructed sets.

We denote MAD(N) = {S↵

: ↵ 2 A} for some set A. Let X := N [MAD(N)with the smallest sequential structure such that

S↵

! {S↵

}, {S↵

} 2 MAD(N).

This condition can be rephrased: members of S↵

as points in N converge to S↵

as point in MAD(N). Then every continuous mapping to R is bounded but thespace is evidently not compact.

To prove this let f : X ! R be a continuous unbounded mapping. That is,there exists a sequence {x

n

} for which f(xn

) ! 1. Now we examine two cases.First, infinitely many members of {x

n

} lie in N, so we can find a subsequence{x

nk} ⇢ N. Then there exists S

↵

2 MAD(N) such that {xnk} ! S

↵

. Then wehave {f(x

nk)} ! f(S

↵

) and {f(xnk)} ! 1 which is a contradiction.

Second, only finite number of members of {xn

} lie in N. Without loss ofgenerality we suppose that {x

n

} = {S↵n} ⇢ MAD(N). For every n 2 N we can

find y

n

2 N such that |f(yn

) � f(S↵n)| 1. The first case yields {f(y

n

)} isbounded and so {f(S

↵n)} is bounded.

14