Factoring out Intuitionistic Theorems:
Continuity Principles and the Uniform
Continuity Theorem
Iris Loeb
Department of Mathematics and Statistics,University of Canterbury,
Private Bag 4800,Christchurch,New Zealand
I.Loeb@math.canterbury.ac.nz
Abstract.
We prove the equivalence between some intuitionistic the
orems and the conjunction of a continuity principle and a compactness
principle over Bishop’s Constructive Mathematics within the programme
of Constructive Reverse Mathematics.To clarify our line of thought,we
ﬁrst point out the relation between quasiequicontinuity,quasiuniform
convergence,and the continuity principle saying that the limit of a con
vergent sequence of continuous functions is again continuous.Finally,as
a spinoﬀ,we conclude that we have found a new,more economic proof of
the statement that every convergent sequence of functions on a compact
metric space converges uniformly.
Keywords:
Constructive Mathematics,Reverse Mathematics,Intuition
istic Mathematics,Continuity Principle,Compactness Principle.
1 Introduction
Intuitionistic mathematics has two clearly separable traits.Firstly,that all func
tions from a complete,separable metric space are continuous (a property that is
also true in the recursive interpretation) [12],and secondly that all continuous
functions on a compact metric space are uniformly continuous (a property that
is also classically valid).
This cognitive distinction can be found back in the results obtained within
the programme of Constructive Reverse Mathematics [13].Many theorems have
been proved equivalent over Bishop’s constructive mathematics (BISH,[7]) to
either a continuity principle [12] or a compactness principle [14,3,4,6,5,8,10] but
not much is known about the interplay between principles of these two classes.
In this paper we will study convergent sequences of functions and we will
single out theorems that are equivalent to the conjunction of a continuity and a
compactness principle over BISH
1
.This contributes not only to the programme
1
However,we do not adopt Bishop’s terminology:for our purposes it is important to
make the distinction between continuity and uniform continuity.
of Constructive Reverse Mathematics,but also to our understanding of intuition
istic mathematics in itself:One of our results is a new intuitionistic proof of the
statement that every pointwise convergent sequence of realvalued (continuous)
functions on a compact metric space converges uniformly.We will compare the
principles used in the new proof with the ones used in [15] and conclude that
the new proof is more economic.
Before giving the main results (Sections 3 and 4),we will ﬁrst take a closer
look at the continuity principles and compactness principle that play an impor
tant role here (Section 2).
2 The Continuity and Compactness Principles
The continuity principles that we consider are the Continuity Principle for Lim
its on Compact Spaces (CPL
cp
) and the Continuity Principle for Compact
Spaces (CONT
cp
).
CPL
cp
:Every pointwise convergent sequence of continuous functions
from a compact metric space to a metric space has a continuous limit.
CONT
cp
:Every function froma compact metric space to a metric space
is continuous.
To our knowledge,these principles have not been examined before,although
the principles CONT
c
(a continuity principle for complete spaces) and CONT
cs
(a continuity principle for complete,separable spaces),which imply CONT
cp
,
were dealt with in [12].Remark that it thus follows from that paper that
CONT
cp
and CPL
cp
are true intuitionistically.
Furthermore,we will use the compactness principle Uniform Continuity The
orem (UCT,[8,10]),
UCT:Every continuous function froma compact metric space to a met
ric space is uniformly continuous.
Because the principle CPL
cp
has not been studied before,we will now show
some important equivalents of it that are wellknown to hold classically [1,2].
This also connects more clearly the results in this paper with the ones of [10],
where we investigated (uniform) equicontinuity and (uniform) convergence.
Let X and Y be metric spaces.A sequence (f
n
)
n≥0
of functions from X to
Y is called quasiequicontinuous if for each > 0 and each x ∈ X we can ﬁnd
δ > 0 such that for each y ∈ X there exists n ∈ N with the property that for all
m≥ n:
If ρ(x,y) < δ,then ρ(f
m
(x),f
m
(y)) < .
Clearly the notion of quasiequicontinuity on a compact metric space X is
weaker than that of equicontinuity on X,as not all terms of a quasiequicontinuous
sequence need be continuous.Even if we require all terms of a quasiequicontinuous
sequence to be continuous,this will not be enough make the sequence equicon
tinuous,as we show in Theorem 3.
The next theorem shows a connection between converging sequences having
a continuous limit and being quasiequicontinuous.Note that we do not require
that the terms of the sequence are continuous.
Theorem 1.
Let X and Y be metric spaces,and let (f
n
)
n≥0
be a sequence of
functions from X to Y that converges to a limit f.Then the following statements
are equivalent.
(i)
The limit function f is continuous on X.
(ii)
The sequence (f
n
)
n≥1
is quasiequicontinuous.
Proof.
Assume (i),let > 0 and x ∈ X.Determine δ > 0 such that if ρ(x,y) < δ,
then ρ(f(x),f(y)) < /3.Consider y ∈ X such that ρ(x,y) < δ.We can ﬁnd
n
0
,n
1
such that ρ(f
m
(x),f(x)) < /3 for all m > n
0
,and ρ(f
m
(y),f(y)) < /3
for all m> n
1
.Setting
N:= max{n
0
,n
1
},
we see that if m> N,then
ρ(f
m
(x),f
m
(y)) ≤ ρ(f
m
(x),f(x)) +ρ(f(x),f(y)) +ρ(f(y),f
m
(y))
<
3
+
3
+
3
= .
Thus (i) implies (ii).
For the reverse implication,assume (ii),and again let > 0 and x ∈ X.
Determine δ > 0 such that for each y ∈ X there exists n such that if ρ(x,y) < δ,
then ρ(f
m
(x),f
m
(y)) < /3 for all m > n.Consider such a point y and the
corresponding n.There exist n
0
,n
1
such that ρ(f
m
(x),f(x)) < /3 for all m >
n
0
,and ρ(f
m
(y),f(y)) < /3 for all m> n
1
.Setting
N:= max{n,n
0
,n
1
},
we see that if m> N,then
ρ(f(x),f(y)) ≤ ρ(f(x),f
m
(x)) +ρ(f
m
(x),f
m
(y)) +ρ(f
m
(y),f(y))
<
3
+
3
+
3
= .
Thus (ii) implies (i).
Let X and Y be metric spaces,and let (f)
n≥0
be a sequence of functions
from X to Y converging pointwise to a limit f.We say that the sequence is
quasiuniformly convergent if for each > 0 and each x ∈ X we can ﬁnd
δ > 0 and n ∈ N with the property that for each y ∈ X:
If ρ(x,y) < δ,then ρ(f
n
(y),f(y)) < .
Usually the deﬁnition of quasiuniformly convergence is formulated in the
following way (see for example [11]),which we will refer to as ﬁnitecover
quasiuniform convergence:Let X and Y be metric spaces,and let (f)
n≥0
be a sequence of functions from X to Y converging pointwise to a limit f.We
say that the sequence is ﬁnitecover quasiuniformly convergent if for each
> 0,there exists a ﬁnite family {V
0
,V
1
,...,V
k
} of open sets of X and a ﬁnite
family {n
0
,n
1
,...,n
k
} of natural numbers
2
,such that
X = V
0
∪V
1
∪...∪V
k
,(1)
and
V
i
⊆ {x ∈ X:ρ(f
n
i
(x),f(x)) < } (i = 0,1,...,k).(2)
The next theorem shows that the notions of quasiuniform convergence and
ﬁnitecover quasiuniform convergence coincide on compact spaces in both clas
sical and intuitionistic mathematics.
Let the Principle of Choice for Compact Spaces be the following state
ment:
For every compact metric space X,if P ⊆ X ×N,and for each x ∈ X
there exists n ∈ N such that (x,n) ∈ P,then there is a function f:X →
N such that (x,f(x)) ∈ P for all x ∈ X.
Note that this principle follows from the Principle of Continuous Choice,and
is hence intuitionistically valid.The Axiom of Choice ensures that it is also
classically valid.
Theorem 2.
Let X be compact metric space,Y a metric space,and (f
n
)
n≥0
a
sequence of continuous functions from X to Y that converges to a limit f.Then
the following statements are equivalent under the assumption of the HeineBorel
Covering Lemma and the Principle of Choice for Compact Spaces.
(i)
The sequence (f
n
)
n≥0
converges quasiuniformly.
(ii)
The sequence (f
n
)
n≥0
converges ﬁnite cover quasiuniformly.
Proof.
Supposing that (i) holds,let > 0.For every x ∈ X determine (with the
Principle of Choice for Compact Spaces) m
x
,n
x
∈ N such that for all y ∈ X
ρ(x,y) < 2
−m
x
⇒ρ(f
n
x
(y),f(y)) < .
Deﬁne for every x ∈ X
V
x
:= {y ∈ X:ρ(x,y) < 2
−m
x
}.
Suppose that y ∈ V
x
for a certain x ∈ X.Then ρ(x,y) < 2
−m
x
and thus
y ∈ {z ∈ X:ρ(f
n
x
(z),f(z)) < }.
2
We remark that [11] requires additionally that these numbers be arbitrarily large.
This requirement can be omitted [9].
This means that if (V
x
)
x∈X
has a ﬁnite subcover,then (2) holds.We get a ﬁnite
subcover by the HeineBorel Covering Lemma,because (V
x
)
x∈X
is indeed a cover
of X.Note that (1) holds by the deﬁnition of V
x
.Hence (i) implies (ii).
To prove the converse,assume (ii),let > 0 and x ∈ X.With V
0
,...,V
k
as in the deﬁnition of “ﬁnitecover quasiuniform convergence”,pick i such that
x ∈ V
i
.Because V
i
is open,we can ﬁnd δ > 0 such that
{y ∈ X:ρ(x,y) < δ} ⊆ V
i
.
Let y ∈ X be such that ρ(x,y) < δ.Then y ∈ V
i
and thus ρ(f
n
i
(y),f(y)) < .
In the rest of the paper we will only use quasiuniformconvergence.The next
theorem connects the property of having a continuous limit to the property of
being quasiuniformly convergent.This also shows that for sequences of continu
ous functions,quasiuniform convergence (resp.,quasiequicontinuity) is strictly
weaker than uniform convergence (resp.,equicontinuity),since classically there
are convergent sequences of continuous functions on [0,1] with a continuous limit
that are not uniformly convergent (resp.,equicontinuous).
Theorem 3.
Let X and Y be metric space,and let (f
n
)
n≥0
be a sequence of
continuous functions from X to Y that converges to a limit f.Then the following
statements are equivalent.
(i)
The limit function f is continuous.
(ii)
The sequence (f
n
)
n≥0
converges quasiuniformly.
Proof.
Assuming (i),let > 0 and x ∈ X.Determine δ
0
> 0 such that for all
y ∈ X if ρ(x,y) < δ
0
then ρ(f(x),f(y)) < /3.Also determine n such that
ρ(f
m
(x),f(x)) < /3 for all m ≥ n.Using the continuity of f
n
,we can ﬁnd
δ
1
> 0 such that for all y ∈ X if ρ(x,y) < δ
1
,then ρ(f
n
(x),f
n
(y)) < /3.Let
δ:= min{δ
0
,δ
1
},
and consider y ∈ X such that ρ(x,y) < δ.We have
ρ(f
n
(y),f(y)) ≤ ρ(f
n
(y),f
n
(x)) +ρ(f
n
(x),f(x)) +ρ(f(x),f(y))
<
3
+
3
+
3
= .
Thus (i) implies (ii).
Now assume (ii),and again let > 0 and x ∈ X.Determine δ
0
> 0 and
n ∈ N such that for each y ∈ X,if ρ(x,y) < δ
0
,then ρ(f
n
(y),f(y)) < /3.By
the continuity of f
n
,there exists δ
1
> 0 such that for all y ∈ X,if ρ(x,y) < δ
1
,
then ρ(f
n
(x),f
n
(y)) < /3.Setting
δ:= min{δ
0
,δ
1
},
consider y ∈ X such that ρ(x,y) < δ.We have
ρ(f(x),f(y)) ≤ ρ(f(x),f
n
(x)) +ρ(f
n
(x),f
n
(y)) +ρ(f
n
(y),f(y))
<
3
+
3
+
3
= .
This completes the proof that (ii) implies (i).
Corollary 1.
The following are equivalent over BISH.
(i)
CPL
cp
.
(ii)
Every convergent sequence of continuous functions on a compact metric space
is quasiequicontinuous.
(iii)
Every convergent sequence of continuous functions on a compact metric space
is quasiuniformly convergent.
Proof.
Immediate,by Theorems 1 and 3.
3 Equivalents of UCT∧CPL
cp
In this section and the next we present some equivalents of a UCT and a conti
nuity principle.These theorems are a strengthening of results in [10].There we
proved the following equivalences:
Theorem 4.
The following statements are equivalent over BISH:
(i)
UCT.
(ii)
For each equicontinuous sequence (f
n
)
n0
of realvalued continuous functions
on [0,1],if {f
i
(x):i ∈ N} is totally bounded for every x ∈ [0,1],then (f
n
)
n0
is uniformly equicontinuous.
(iii)
Every equicontinuous,convergent sequence of realvalued continuous func
tions on [0,1] has a uniformly continuous limit.
(iv)
Every equicontinuous,convergent sequence of realvalued continuous func
tions on [0,1] is uniformly convergent.
This will be used to obtain equivalences to UCT∧CPL
cp
.If we compare The
orem 4 with Theorem 5,we see illustrated how CPL
cp
in the latter replaces the
requirement of equicontinuity in the former.
Theorem 5.
The following statements are equivalent over BISH:
(i)
UCT∧CPL
cp
.
(ii)
For every compact metric space X:for each sequence (f
n
)
n≥0
of continuous
functions from X to a metric space,if {f
i
(x):i ∈ N} is totally bounded for
every x ∈ X,then (f
n
)
n≥0
is uniformly equicontinuous.
(iii)
Every convergent sequence of continuous functions from a compact metric
space to a metric space has a uniformly continuous limit.
(iv)
Every convergent sequence of continuous functions from a compact metric
space to a metric space is uniformly convergent.
Before giving the proof of Theorem 5,we generalise two results of [10].
Lemma 1.
Let X,Y be two metric spaces.Then every uniformly equicontinu
ous,convergent sequence of functions from X to Y has a uniformly continuous
limit.
Proof.
Let X,Y be two metric spaces,and let (f
n
)
n≥0
be a uniformly equicon
tinuous sequence of functions from X to Y that converges to a limit f.Given
> 0,determine δ such that for all i ∈ N and all x,y ∈ X,if ρ(x,y) < δ,then
ρ(f
i
(x),f
i
(y)) <
1
3
.Let x,y ∈ X such that ρ(x,y) < δ.Determine m
0
,m
1
such
that ρ(f
n
(x),f(x)) <
1
3
for all n ≥ m
0
,and ρ(f
n
(y),f(y)) <
1
3
for all n ≥ m
0
.
Deﬁne
p = max{m
0
,m
1
}.
Then
ρ(f(x),f(y)) ≤ ρ(f
p
(x),f(x)) +ρ(f
p
(y),f(y)) +ρ(f
p
(x),f
p
(y))
<
1
3
+
1
3
+
1
3
= .
Lemma 2.
Let X be a totally bounded metric space,Y a metric space,and
(f
n
)
n≥0
a convergent sequence of functions from X to Y.If (f
n
)
n≥0
is uniformly
equicontinuous,then (f
n
)
n≥0
converges uniformly.
Proof.
Let X be a totally bounded metric space,Y a metric space,and (f
n
)
n≥0
a sequence of functions from X to Y that converges to a limit f.Suppose that
(f
n
)
n≥0
is uniformly equicontinuous.Then f is uniformly continuous (by Lemma
1).Given > 0,determine δ
0
and δ
1
such that for all x,y ∈ X and each
i ∈ N,if ρ(x,y) < δ
0
,then ρ(f
i
(x),f
i
(y)) <
1
3
,and if ρ(x,y) < δ
1
,then
ρ(f(x),f(y)) <
1
3
.Determine k such that 2
−k
< min{δ
0
,δ
1
}.Let A be a 2
−k

approximation of X.For every j ∈ A pick n
j
such that
ρ(f(j),f
m
(j)) <
1
3
for every m> n
j
.Deﬁne
N:= max{n
j
:j ∈ A}.
Let r ≥ N and x ∈ X.Determine p ∈ A such that ρ(p,x) < 2
−k
.Then
ρ(f(x),f
r
(x)) ≤ ρ(f(p),f(x)) +ρ(f
r
(p),f
r
(x)) +ρ(f
r
(p),f(p))
<
1
3
+
1
3
+
1
3
=
Now we give the proof of Theorem 5.
Proof
(of Theorem 5).We will ﬁrst show that (i) implies (ii).Assume (i);let X
be a compact metric space;let Y be a metric space;let (f
n
)
n≥0
be a sequence of
continuous functions from X to Y such that {f
i
(x):i ∈ N} is totally bounded
for every x ∈ X.Then
{ρ(f
i
(x),f
i
(y)):i ∈ N}
is totally bounded as well.Hence
sup{ρ(f
i
(x),f
i
(y)):i ∈ N}
exists for every x,y ∈ X.Deﬁne a sequence (g
n
)
n≥0
:X ×X →R by:
g
n
(x,y):= max
i≤n
ρ(f
i
(x),f
i
(y))
Then g
n
is continuous for every n.Moreover for every x,y ∈ X,the sequence
(g
n
(x,y))
n≥0
converges to
g(x,y):= sup{ρ(f
i
(x),f
i
(y)):i ∈ N}.
Therefore,by CPL
cp
,g is continuous.So g is uniformly continuous by UCT.Let
> 0.Determine δ such that for all (x,y),(x
,y
) ∈ X×X,if ρ((x,y),(x
,y
)) <
δ,then g(x,y) −g(x
,y
) < .Let x,y ∈ X be such that ρ(x,y) < δ,and let
i ∈ N.Then
ρ(f
i
(x),f
i
(y)) ≤ sup{ρ(f
j
(x),f
j
(y)):j ∈ N}
= g(x,y) = g(x,y) −g(x,x) <
Because it is immediately clear that (iii) follows from (i),we continue with
the proof of (iv) from (ii).So assume (ii).Let X be a compact metric space,let
Y be a metric space,and let (f
n
)
n0
be a convergent sequence (with a limit f)
of continuous functions fromX to Y.Note that,because the sequence converges,
{f
i
(x):i ∈ N} is totally bounded for every x ∈ X.Hence,by statement (ii),
the sequence (f
n
)
n0
is uniformly equicontinuous.So we conlclude by Lemma 2
that (f
n
)
n0
converges uniformly.
We now prove the reverse direction.It follows directly from Theorem 4 that
the statements (ii),(iii),and (iv) imply UCT.Remark that it is also immediately
clear that each of the statements (iii) and (iv) implies CPL
cp
.To prove that
(ii) implies CPL
cp
,let X be a compact metric space,let Y be a metric space,
and let (f
n
)
n0
be a sequence of continuous functions from X to Y converging
to a limit f.Then {f
i
(x):i ∈ N} is totally bounded for every x ∈ X.Thus,by
(ii),the sequence is uniformly equicontinuous.Therefore f is continuous.
This completes the proof.
Note that the theorem
Every convergent quasiequicontinuous sequence of continuous functions
from a compact metric space to a metric space has a uniformly continu
ous limit.
is equivalent to UCT,but that neither of the following two statements is so
equivalent:
For every compact metric space X:for each quasiequicontinuous se
quence (f
n
)
n≥0
of continuous functions from X to a metric space,if
{f
i
(x):i ∈ N} is totally bounded for every x ∈ X,then (f
n
)
n≥0
is
uniformly equicontinuous.
Every quasiequicontinuous convergent sequence of continuous functions
from a compact metric space to a metric space is uniformly convergent.
4 Equivalents of UCT∧CONT
cp
The next theorem follows naturally from the results of the previous section.
Theorem 6.
The following statements are equivalent over BISH:
(i)
UCT∧CONT
cp
.
(ii)
For every compact metric space X:for each sequence (f
n
)
n≥0
of functions
from X to a metric space,if {f
i
(x):i ∈ N} is totally bounded for every
x ∈ X,then (f
n
)
n≥0
is uniformly equicontinuous.
(iii)
Every convergent sequence of functions from a compact metric space to a
metric space has a uniformly continuous limit.
Proof.
The fact that (i) implies the statements (ii),and (iii) is an immediate con
sequence of Theorem5.That these statements imply UCT follows also fromthe
results in the previous section.We will now show that they also imply CONT
cp
.
Let X be a compact metric space,Y a metric space,and let f be a function
from X to Y.Deﬁne a sequence (f
n
)
n≥0
by f
n
= f for every n ∈ N.Observe
that this sequence converges.Then the conclusions that the sequence is uni
formly equicontinuous or that f is uniformly continuous,both imply that f is
continuous.
We have only a partial generalisation of Theorem 5 item (iv) in the sense of
Theorem 6,which is an immediate consequence of the former theorem.
Theorem 7.
UCT∧ CONT
cp
implies over BISH that every convergent se
quence of functions from a compact metric space to a metric space is uniformly
convergent.
Let us now come back to the original proof of the statement that every
pointwise convergent sequence of realvalued functions on a compact metric space
converges uniformly [15].Without stating explicitly that it does so,this proof
uses CONT
cp
and the Full Fan Theorem.
Our results do not show whether we can weaken the continuity assumption
without strengthening another hypothesis.However,if we assume that the func
tions of the sequence are continuous,then a weaker continuity principle suﬃces
(Theorem 5,(iv)).
On the other hand,it is possible to economise on the strength of the compact
ness principle.Instead of using the Full Fan Theorem,we may assume just UCT
(Theorem 7),which is believed to be weaker [10],even under the assumption of
CONT
cp
.Note that we could not have concluded this directly from Theorem 4,
as we do not have a proof that CONT
cp
implies that all convergent sequences
of functions on a compact space are equicontinuous.
Acknowledgments.The author thanks the Marsden Fund of the Royal Soci
ety of New Zealand for supporting her by a Postdoctoral Research Fellowship.
She also thanks Douglas Bridges,Hannes Diener and Josef Berger for useful
discussions,and the anonymous reviewers for useful comments.
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