Limit everywhere => Continuous almost everywhere?
Guy Ritchie
suppose f : [a,b] -> R is a function such that its limit exists everywhere.
i.e g: [a,b] -> R such that g(x) = Limit_{t -> x} f(t) is well defined.
We can easily show that g is a continuous function. But what about f?
Can we show that it is continous almost everywhere?
[a,b] is the closed interval {x | a <= x <= b} and R is the set of reals.
Arturo Magidin
Yes. f has at most finitely many discontinuities, if I am not
mistaken (my proof below uses the Axiom of Choice):
Suppose f has infinitely many distinct discontinuities. Extract a
contable set, and choose a bijection with N, so we have a sequence
x_1,...,x_n,... of distinct discontinuities.
Since we are in a closed interval [a,b], hence a compact set, there
must be a converging subsequence ; replace (x_i) by that sequence if
necessary, so we may assume that x_1,...,x_n converges to a point x in
[a,b], and all x_i are distinct. If necessary, remove x from the
sequence to get a converging sequence of distinct points, which
converges to a point not in the sequence.
Let e>0. Then there exists d>0 such that for every y in [a,b], if
0<|x-y|<d then |g(x)-f(y)|<e (since the limit at x exists, and is
equal to g(x) ). Therefore, if y,z in [a,b] satisfy 0<|x-y|<d and
0<|x-z|<d, then |f(z)-f(y)|<= |f(z)-g(x)|+|g(x)-f(y)| < 2e.
Pick N(d) such that for all n>N(d), 0<|x_n-x|<d/2 (the 'greater than
zero' is possible since x is not equal to x_n for all n).
Since f is discontinuous at each x_i, for each n>N(d) there exists
z(n) in [a,b] such that |z-x_n|<|x_n-x|/2, but such that
|f(z)-f(x_n)|>2e. In particular, note that z is not equal to x, and
that |z-x_n| < (d/4).
Now consider z(n) and x_n. We know that 0<|x_n-x|<d/2; also, that
0<|z-x| = |z-x_n + x_n - x| <= |z-x_n| + |x_n-x| < (d/4)+(d/2) < d.
Therefore, by what we said above, we must have that
|f(z)-f(x_n)|<2e
but this contradicts our choice of z.
This contradiction arises from the assumption that f has infinitely
many distinct discontinuities in [a,b]; hence, f can have at most
finitely many discontinuities. In the usual measure, any finite set
has measure zero, so f is continuous almost everywhere.
Unless I made a mistake somewhere...
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
I think so; I think f will have at most countably many
discontinuities. (I posted an argument with a very stupid error
before, supposedly proving only finitely many discontinuities; sorry
about that. I cancelled that post. I realized the stupid error halfway
home, and I actually drove back to the office to correct it...)
Let e>0. For each x in [a,b] there exists d(x)>0 such that for all y
in [a,b], if 0<|x-y|<d(x), then |g(x)-f(y)|<e/2, since the limit
always exists and is equal to g(x). Therefore, for every x in [a,b],
there exists d(x)>0 such that for all y,z in [a,b], if 0<|x-y|<d(x)
and 0<|x-z|<d(x), then |f(z)-f(y)|<= |f(z)-g(x)| + |g(x)-f(y)| < e.
Consider the open cover of intervals of radius d(x)/2 around x. Since
[a,b] is compact, there is a finite subcover, consisting of balls
around x_i with radius d(x_i)/2, i=1,...,n.
Let d be the smallest of d(x_1),...,d(x_n).
Let y and z be any two points in [a,b] such that:
(1) y and z are not equal to any of x_1,...,x_n
(2) |y-z|< d/2.
Since the balls cover the interval, there must exist x_i such that
|y-x_i|<d(x_i)/2 < d(x_i)
But then
|z-x_i| <= |z-y| + |y-x_i| < d/2+d(x_i)/2 < d/2+d/2 = d < d(x_i)
But then neither z nor y are equal to x_i, and both are within d(x_i)
of x, so from the above we know that
|f(z)-f(y)|<e.
For, for every e>0, there exists a finite set of points x_1,...,x_n,
and a positive integer d>0 such that, if y and z are any two points in
[a,b], none of them equal to any of {x1,...,xn}, and |y-z|<d, then
|f(z)-f(y)|<e.
It looks like we're done, but we are not really, because the points
x_1,...,x_n depend on e. Let's call those points "possible problem
points for e".
Now consider the sequence of numbers 1/n, with n a positive integer.
Let P={x_1,...,x_m,...} be the union of the collections of possible
problem points for 1/n, as n ranges over the positive integers. This
is a countable set, since it is a countable union of finite sets.
I claim that f is continuous at every point in [a,b]-P.
Let x be a point not in P, and let e>0. We want to show that there
exists d>0 such that if |y-x|<d, and y is in [a,b], then |f(y)-f(x)|<e.
Let n be a positive integer such that 1/n < e.
Then there is a d(n)>0 such that no possible problem point for 1/n lies
within d(n) of x, as there are only finitely many possible problem
points for 1/n.
And there is a d'>0 such that, for any y and z which are not possible
problem points for 1/n, and such that |y-z|<d', we have
|f(z)-f(y)|<e, as seen above.
Let d be the smallest of d(n) and d'. Then for all y in [a,b], with
|y-x|<d, we must have:
(1) That y and x are not possible problem points for 1/n, since they are
within d(n) of x;
and
(2) |y-x|<d'.
Therefore, |f(x)-f(y)|<e.
In summary, if y is in [a,b] and |y-x|<d, then |f(x)-f(y)|<e.
This proves that f is continuous at x.
Therefore, the only possible points of discontinuity are the possible
problem points for 1/n, n=1,...,infty. This is a countable set. In the
usual measure for the real numbers, countable sets are of measure
zero, so f(x) must be continuous outside a countable set, hence
continuous almost everywhere.
NOTE: I ->think<- it may be true that f(x) must be continuous in all
but finitely many points. Does anybody know if this is true?
(Of course, it is possible that by "almost everywhere" you meant "at
all except possibly for a finite number of points", in which case I'm
not done; but 'almost everywhere' usually means 'outside a set of
measure zero', so hopefully this does it).
I hope I didn't make another silly mistake here...
Arturo Magidin
I think so; I think f will have at most countably many
discontinuities. (I posted an argument with a very stupid error
before, supposedly proving only finitely many discontinuities; sorry
about that. I cancelled that post. I realized the stupid error halfway
home, and I actually drove back to the office to correct it...)
Let e>0. For each x in [a,b] there exists d(x)>0 such that for all y
in [a,b], if 0<|x-y|<d(x), then |g(x)-f(y)|<e/2, since the limit always
exists. Therefore, for every x in [a,b], there exists d(x)>0 such that
for all y,z in [a,b], if 0<|x-y|<d(x) and 0<|x-z|<d(x), then
|f(z)-f(y)|<= |f(z)-g(x)| + |g(x)-f(y)| < e.
Consider the open cover of intervals of radius d(x)/2 around x. Since
[a,b] is compact, there is a finite subcover, consisting of balls
around x_i with radius d(x_i)/2, i=1,...,n.
Let d be the smallest of d(x_1),...,d(x_n).
Let y and z be any two points in [a,b] such that:
(1) y and z are not equal to any of x_1,...,x_n
(2) |y-z|< d/2.
Since the balls cover the interval, there must exist x_i such that
|y-x_i|<d(x_i)/2 < d(x_i)
But then
|z-x_i| <= |z-y| + |y-x_i| < d/2+d(x_i)/2 <= d(x_i)/2+d(x_i)/2 = d(x_i)
So then neither z nor y are equal to x_i, and both are within d(x_i)
of x, so from the above we know that
|f(z)-f(y)|<e.
IN summary, for every e>0 there exists a finite set x1,...,xn of
points in [a,b], and a d>0 such that if y and z are any two points
in [a,b] which are not equal to any x1,...,xn, and |y-z|<d, then
|f(y)-f(z)|<e.
It looks like we're done, but we are not really, because the points
x_1,...,x_n depend on e. Let's call those points "possible problem
points for e".
Now consider the sequence of numbers 1/n, with n a positive integer.
Let P={x_1,...,x_m,...} be the union of the collections of possible
problem points for 1/n, as n ranges over the positive integers. This
is a countable set, since it is a countable union of finite sets.
I claim that f is continuous at every point in [a,b]-P.
Let x be a point not in P, and let e>0. We want to show that there
exists d>0 such that if |y-x|<d, and y is in [a,b], then |f(y)-f(x)|<e.
Let n be a positive integer such that 1/n < e.
Then there is a d(n)>0 such that no possible problem point for 1/n lies
within d(n) of x, as there are only finitely many possible problem
points for 1/n.
And there is a d'>0 such that, for any y and z which are not possible
problem points for 1/n, and such that |y-z|<d', we have
|f(z)-f(y)|<e, by the above.
Let d be the smallest of d(n) and d'. Then for all y in [a,b], with
|y-x|<d, we must have:
(1) That y and x are not possible problem points for 1/n, since they are
within d(n) of x;
and
(2) |y-x|<d'.
Therefore, |f(x)-f(y)|<e.
In summary, if y is in [a,b] and |y-x|<d, then |f(x)-f(y)|<e.
This proves that f is continuous at x.
Therefore, the only possible points of discontinuity are the possible
problem points for 1/n, n=1,...,infty. This is a countable set. In the
usual measure for the real numbers, countable sets are of measure
zero, so f(x) must be continuous outside a countable set, hence
continuous almost everywhere.
NOTE: I ->think<- it may be true that f(x) must be continuous in all
but finitely many points. Does anybody know if this is true? Perhaps
some sort of diagonal argument and compactness?
I hope I dind't make any silly mistakes this time...
I guess it's possible that by "almost everywhere" you meant at all but
a finite number of points, in which case I wouldn't be done yet...
Fred Galvin
> In article <66070603.02111...@posting.google.com>,
> Guy Ritchie <googlep...@yahoo.com> wrote:
> >
> >suppose f : [a,b] -> R is a function such that its limit exists everywhere.
> >i.e g: [a,b] -> R such that g(x) = Limit_{t -> x} f(t) is well defined.
> >
> >We can easily show that g is a continuous function. But what about f?
> >Can we show that it is continous almost everywhere?
> >
> >[a,b] is the closed interval {x | a <= x <= b} and R is the set of reals.
>
> I think so; I think f will have at most countably many
> discontinuities. [. . .]
Right. No doubt your argument (which I haven't read) is correct, but
it seems rather long. Let e be any positive number ("not necessarily
the base of natural logarithms"). For each point x_0, we can choose a
neighborhood U of x_0 such that |f(x)-g(x_0)| < e/2 and |g(x)-g(x_0)|
< e/2 for all x in U\{x_0}, whence |f(x)-g(x)| < e for all but one
point x in U. The domain space [a,b] is covered with such U's. Taking
a countable (in case somebody wants to generalize this to abstract
topological spaces, only the Lindelof property is used here) subcover,
we see that |f(x)-g(x)| < e except at countably many points. Doing
this for e = 1/n, n = 1, 2, 3, ..., we see that f(x) = g(x) for all
but countably many x. Of course, if f(x) = g(x), then f is continuous
at x. (I guess your argument must be the same, in a more verbose
style.)
> NOTE: I ->think<- it may be true that f(x) must be continuous in all
> but finitely many points. Does anybody know if this is true?
No. Let [a,b] = [0,1], define f(1/n) = 1/n for n = 1, 2, 3, ..., and
define f(x) = 0 otherwise.
More interestingly, f can have a dense set of discontinuities. Namely,
let f(x) = 0 if x is irrational, f(m/n) = 1/|n| if m/n is a reduced
fraction. This is the classical example of a function which is
discontinuous at each rational point on the line, and continuous at
each irrational point; it has limit 0 at every point.
The World Wide Wade
Fred Galvin <gal...@math.ukans.edu> wrote:
> On Sat, 16 Nov 2002, Arturo Magidin wrote:
>
> > In article <66070603.02111...@posting.google.com>,
> > Guy Ritchie <googlep...@yahoo.com> wrote:
> > >
> > >suppose f : [a,b] -> R is a function such that its limit exists everywhere.
> > >i.e g: [a,b] -> R such that g(x) = Limit_{t -> x} f(t) is well defined.
> > >
> > >We can easily show that g is a continuous function. But what about f?
> > >Can we show that it is continous almost everywhere?
> > >
> > >[a,b] is the closed interval {x | a <= x <= b} and R is the set of reals.
> >
> > I think so; I think f will have at most countably many
> > discontinuities. [. . .]
>
> Right. No doubt your argument (which I haven't read) is correct, but
> it seems rather long. Let e be any positive number ("not necessarily
> the base of natural logarithms"). For each point x_0, we can choose a
> neighborhood U of x_0 such that |f(x)-g(x_0)| < e/2 and |g(x)-g(x_0)|
> < e/2 for all x in U\{x_0}, whence |f(x)-g(x)| < e for all but one
> point x in U. The domain space [a,b] is covered with such U's. Taking
> a countable (in case somebody wants to generalize this to abstract
> topological spaces, only the Lindelof property is used here) subcover,
> we see that |f(x)-g(x)| < e except at countably many points. Doing
> this for e = 1/n, n = 1, 2, 3, ..., we see that f(x) = g(x) for all
> but countably many x. Of course, if f(x) = g(x), then f is continuous
> at x. (I guess your argument must be the same, in a more verbose
> style.)
Alternatley you can let D_n = {x in [a,b] : |f(x)-g(x)| > 1/n}. The set D
of discontinuities is the union of the D_n's. If D were not countable, then
some D_n would be uncountable, hence infinite, hence have an accumulation
point p somewhere in [a,b]. It's then easy to check that lim_{x->p} f(x)
fails to exist, contradiction.
Guy Ritchie
> On Sat, 16 Nov 2002, Arturo Magidin wrote:
> Right. No doubt your argument (which I haven't read) is correct, but
> it seems rather long. Let e be any positive number ("not necessarily
> the base of natural logarithms"). For each point x_0, we can choose a
> neighborhood U of x_0 such that |f(x)-g(x_0)| < e/2 and |g(x)-g(x_0)|
> < e/2 for all x in U\{x_0}, whence |f(x)-g(x)| < e for all but one
> point x in U. The domain space [a,b] is covered with such U's. Taking
> a countable (in case somebody wants to generalize this to abstract
> topological spaces, only the Lindelof property is used here) subcover,
> we see that |f(x)-g(x)| < e except at countably many points. Doing
> this for e = 1/n, n = 1, 2, 3, ..., we see that f(x) = g(x) for all
> but countably many x. Of course, if f(x) = g(x), then f is continuous
> at x. (I guess your argument must be the same, in a more verbose
> style.)
Arturo's argument makes use of the fact that
a) if U is a (not necessarily countable) collection of open intervals
covering a closed (and bounded) set, then there is a finite
sub-collection of U
which covers the same set.
you are using the fact that
b) there is a countable subcover.
are there spaces where b) holds but not a) ?
please note that i have started learning these terms by picking up on
your responses to my question.. so please feel free to be pedantic.
José Carlos Santos
> Arturo's argument makes use of the fact that
> a) if U is a (not necessarily countable) collection of open intervals
> covering a closed (and bounded) set, then there is a finite
> sub-collection of U
> which covers the same set.
>
> you are using the fact that
> b) there is a countable subcover.
>
> are there spaces where b) holds but not a) ?
>
> please note that i have started learning these terms by picking up on
> your responses to my question.. so please feel free to be pedantic.
I am not sure that I have undestood your question. If what you're asking
is "are there spaces such that every open cover has a countable
subcover and such that not every open cover has a finite subcover?",
then the answer is "yes": take the reals and the cover whose elements
are the intervals of the form ]n , n + 2[ (n an arbitrary integer).
Best regards
Jose Carlos Santos
Arturo Magidin
>
>> In article <66070603.02111...@posting.google.com>,
>> Guy Ritchie <googlep...@yahoo.com> wrote:
>> >
>> >suppose f : [a,b] -> R is a function such that its limit exists everywhere.
>> >i.e g: [a,b] -> R such that g(x) = Limit_{t -> x} f(t) is well defined.
>> >
>> >We can easily show that g is a continuous function. But what about f?
>> >Can we show that it is continous almost everywhere?
>> >
>> >[a,b] is the closed interval {x | a <= x <= b} and R is the set of reals.
>>
>> I think so; I think f will have at most countably many
>> discontinuities. [. . .]
>
>Right. No doubt your argument (which I haven't read) is correct, but
>it seems rather long. Let e be any positive number ("not necessarily
>the base of natural logarithms"). For each point x_0, we can choose a
>neighborhood U of x_0 such that |f(x)-g(x_0)| < e/2 and |g(x)-g(x_0)|
>< e/2 for all x in U\{x_0}, whence |f(x)-g(x)| < e for all but one
>point x in U. The domain space [a,b] is covered with such U's. Taking
>a countable (in case somebody wants to generalize this to abstract
>topological spaces, only the Lindelof property is used here) subcover,
>we see that |f(x)-g(x)| < e except at countably many points. Doing
>this for e = 1/n, n = 1, 2, 3, ..., we see that f(x) = g(x) for all
>but countably many x. Of course, if f(x) = g(x), then f is continuous
>at x. (I guess your argument must be the same, in a more verbose
>style.)
Heh. That ->is<- essentially the argument I posted. Except that
you can do a bit better and show that |f(x)-g(x)|<e except at finitely
many points for each e.
(My argument was more verbose, because after saying something stupid
in a previous post, I went into overly-cautious-verbose mode to make
sure I didn't say anything stupid again; of course, I sort of did below:
>> NOTE: I ->think<- it may be true that f(x) must be continuous in all
>> but finitely many points. Does anybody know if this is true?
>
>No. Let [a,b] = [0,1], define f(1/n) = 1/n for n = 1, 2, 3, ..., and
>define f(x) = 0 otherwise.
Yup. I thought of that exact example this morning. Thanks!
Guy Ritchie
> Guy Ritchie wrote:
>
> I am not sure that I have undestood your question. If what you're asking
> is "are there spaces such that every open cover has a countable
> subcover and such that not every open cover has a finite subcover?",
> then the answer is "yes": take the reals and the cover whose elements
> are the intervals of the form ]n , n + 2[ (n an arbitrary integer).
>
> Best regards
>
> Jose Carlos Santos
I meant to ask:
Is there a space such that there is some closed bounded set C such
that
every open cover of C has a coutable subcover and there exists an open
cover of
C which does not have a finite subcover.
btw,
how do we define a 'bounded' set for general spaces.
also, can you please provide me with the names of some good
introductory books on topology.
Thanks,
Guy.
Fred Galvin
> I meant to ask:
> Is there a space such that there is some closed bounded set C such
> that every open cover of C has a coutable subcover and there exists
> an open cover of C which does not have a finite subcover.
Hilbert space. Let C be an orthonormal basis or a closed ball.
> btw,
> how do we define a 'bounded' set for general spaces.
I don't know.
> also, can you please provide me with the names of some good
> introductory books on topology.
Stephen Willard, _General Topology_
Robert Israel
Guy Ritchie <googlep...@yahoo.com> wrote:
>how do we define a 'bounded' set for general spaces.
How "general"? In a metric space with metric d, it's a set
S such that there is a constant B such that d(x,y) <= B for all
x,y in S. Of course, this depends on the choice of metric; a set
can be bounded in one metric and not in another, even if the metrics
induce the same topology. So "bounded" is not a topological concept.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Stephen J. Herschkorn
>
>
>I meant to ask:
>Is there a space such that there is some closed bounded set C such
>that
>every open cover of C has a coutable subcover and there exists an open
>cover of
>C which does not have a finite subcover.
>
>btw,
>how do we define a 'bounded' set for general spaces.
>also, can you please provide me with the names of some good
>introductory books on topology.
>
Give R the metric d(x,y) = min(|x-y|, 1). R is bounded under this
metric and it is closed. Use the fact that open intervals with rational
endpoints form a base for the topology to convince yourself that every
open cover of R has a countable subcover.
By the way, a topological space where every open covering has a
countable subcovering is called a Lindelöf space. (That's supposed to
be an umlaut-o in case it doesn't post properly.)
A subset A of a metric space is by definition bounded iff its diameter
(= sup {d(x,y) : x, y in A}) is finite.
A subset A of a topological vector space is by definition bounded iff
for every neighborhood U of 0 there exists t > 0 such that A is a
subset of sU for all s > t.
(True or false? If the TVS is metrizable, bounded in the latter sense
implied bounded in the former sense for any corresponding metric.)
Masters, is there a generalization to uniform spaces?
For the topology of metric spaces, you can look at some undergraduate
analysis texts, such as
Goldberg RR, Methods of Real Analysis, and
Rudin W, Principles of Mathematical Analysis,
though it has been discussed in other threads that there is really no
reason to just start out with general point-set topology. One
elementary undergraduate text on the latter is
Moore TO, Elementary General Topology, Prentice Hall, 1964.
A more comprehensive text on the undergraduate/graduate border is
Munkres JR, Topology.
You might also look at
Kelly JL, Topology.
--
Stephen J. Herschkorn hers...@rutcor.rutgers.edu
Guy Ritchie
Guy.
googlep...@yahoo.com (Guy Ritchie) wrote in message news:<66070603.02111...@posting.google.com>...