My Math Forum counter example to a skewed deltaepsilon defns?

 Real Analysis Real Analysis Math Forum

 October 12th, 2009, 08:14 PM #1 Senior Member   Joined: Apr 2009 Posts: 201 Thanks: 0 counter example to a skewed deltaepsilon defns? can you check this please? disprove: for any delta, there exists an epsilon such that 0 < | x-a| < delta --> |f(x) - L | < epsilon this is just the negative of the usual delta-epsilon definition, so I can show how there's at least one case where there isn't an epsilon for ANY delta... for any epsilon, there exists a delta such that |f(x) - L | < epsilon ---> 0 < | x-a| < delta I picked a periodic function like sin(x) or cos(x) where you can find f(x) = f(x1) = f(x2) = f(x3).. et c. so, f of any of these x's - L should be less than epsilon, but if you then take one of those x's |x- some fixed a| is not within delta - since you can pick an a from a different period such that |f(x) - L| < epsilon but x-a is not less than delta \ edit: nevermind, you could just pick delta to be as big as you want, so this wouldn't work thanks
October 13th, 2009, 05:53 AM   #2
Senior Member

Joined: Dec 2008

Posts: 306
Thanks: 0

Re: counter example to a skewed deltaepsilon defns?

Are you trying to show that if this
Quote:
 for any delta, there exists an epsilon such that 0 < | x-a| < delta --> |f(x) - L | < epsilon
is satisfied for some L, then it does not follow that f has limit L at a ?

October 13th, 2009, 07:59 AM   #3
Senior Member

Joined: Apr 2009

Posts: 201
Thanks: 0

Re: counter example to a skewed deltaepsilon defns?

Quote:
Originally Posted by dman315
Are you trying to show that if this
Quote:
 for any delta, there exists an epsilon such that 0 < | x-a| < delta --> |f(x) - L | < epsilon
is satisfied for some L, then it does not follow that f has limit L at a ?

I need to find an example where a limit would exist but this definition wouldn't be a proper definition

thanks

October 13th, 2009, 09:10 AM   #4
Senior Member

Joined: Oct 2007
From: Chicago

Posts: 1,701
Thanks: 3

Re: counter example to a skewed deltaepsilon defns?

Quote:
 Originally Posted by ElMarsh can you check this please? disprove: for any delta, there exists an epsilon such that 0 < | x-a| < delta --> |f(x) - L | < epsilon this is just the negative of the usual delta-epsilon definition, so I can show how there's at least one case where there isn't an epsilon for ANY delta...
This one is trickier than the second because most of the "natural" continuous functions satisfy this... But you are missing what you need to do a little bit: You want to show there's an example where some delta doesn't doesn't have an epsilon.
Try to find a discontinuous function which satisfies this property.

Quote:
 for any epsilon, there exists a delta such that |f(x) - L | < epsilon ---> 0 < | x-a| < delta I picked a periodic function like sin(x) or cos(x) where you can find f(x) = f(x1) = f(x2) = f(x3).. et c. so, f of any of these x's - L should be less than epsilon, but if you then take one of those x's |x- some fixed a| is not within delta - since you can pick an a from a different period such that |f(x) - L| < epsilon but x-a is not less than delta \ edit: nevermind, you could just pick delta to be as big as you want, so this wouldn't work thanks
Actually. For this second case, this works: For any fixed delta (however large), you can find a larger r for which sin(r)=L (so sin(r)-L<epsilon, for any epsilon). You can pick delta as big as you want, but once you pick it, it's fixed, and you can pick an x that is farther away than delta, but has a "good" f(x)

This doesn't work for the first example, because you can choose any epsilon > 2, and it works (all it says is there exists an epsilon-- it says nothing about whether this epsilon is large or small)

October 13th, 2009, 11:50 AM   #5
Senior Member

Joined: Apr 2009

Posts: 201
Thanks: 0

Re: counter example to a skewed deltaepsilon defns?

Quote:
Originally Posted by cknapp
Quote:
 Originally Posted by ElMarsh can you check this please? disprove: for any delta, there exists an epsilon such that 0 < | x-a| < delta --> |f(x) - L | < epsilon this is just the negative of the usual delta-epsilon definition, so I can show how there's at least one case where there isn't an epsilon for ANY delta...
This one is trickier than the second because most of the "natural" continuous functions satisfy this... But you are missing what you need to do a little bit: You want to show there's an example where some delta doesn't doesn't have an epsilon.
Try to find a discontinuous function which satisfies this property.

Quote:
 for any epsilon, there exists a delta such that |f(x) - L | < epsilon ---> 0 < | x-a| < delta I picked a periodic function like sin(x) or cos(x) where you can find f(x) = f(x1) = f(x2) = f(x3).. et c. so, f of any of these x's - L should be less than epsilon, but if you then take one of those x's |x- some fixed a| is not within delta - since you can pick an a from a different period such that |f(x) - L| < epsilon but x-a is not less than delta \ edit: nevermind, you could just pick delta to be as big as you want, so this wouldn't work thanks
Actually. For this second case, this works: For any fixed delta (however large), you can find a larger r for which sin(r)=L (so sin(r)-L<epsilon, for any epsilon). You can pick delta as big as you want, but once you pick it, it's fixed, and you can pick an x that is farther away than delta, but has a "good" f(x)

This doesn't work for the first example, because you can choose any epsilon > 2, and it works (all it says is there exists an epsilon-- it says nothing about whether this epsilon is large or small)
haha right, I see that the second one works. I came up with it last night, but this morning I woke up and thought to myself - what if you picked delta to be the whole real axis (infinity) :O but of course, that's silly you can't do that

I will think more about the first case

thanks for the help! you're always helping me, I appreciate it

October 13th, 2009, 12:03 PM   #6
Senior Member

Joined: Oct 2007
From: Chicago

Posts: 1,701
Thanks: 3

Re: counter example to a skewed deltaepsilon defns?

Quote:
 Originally Posted by ElMarsh thanks for the help! you're always helping me, I appreciate it
Not a problem. My analysis needs serious work, so seeing these problems is helpful for me as well...

 October 14th, 2009, 05:27 PM #7 Senior Member   Joined: Apr 2009 Posts: 201 Thanks: 0 Re: counter example to a skewed deltaepsilon defns? Hi, I think for the first part, if you have a discontinuous function that has very large intervals between each point or segment, you wouldn't be able to find an epsilon for a certain delta because the epsilon would have to span infinitely
October 15th, 2009, 03:36 AM   #8
Senior Member

Joined: Oct 2007
From: Chicago

Posts: 1,701
Thanks: 3

Re: counter example to a skewed deltaepsilon defns?

Quote:
 Originally Posted by ElMarsh Hi, I think for the first part, if you have a discontinuous function that has very large intervals between each point or segment, you wouldn't be able to find an epsilon for a certain delta because the epsilon would have to span infinitely
The first problem with this is idea is that it doesn't work-- Those gaps need to be well-defined and fixed. For any given delta around a certain point, you will be able to find a "good" epsilon-- it will be very large, no doubt, but it will work.

That doesn't matter though: My suggestion was to look at a discontinuous function which satisfies the statement you gave. If the function is discontinuous, and doesn't satisfy the sentence, you get nothing, because you are trying to show that this definition doesn't line up with continuity. In other words, call
Quote:
 Originally Posted by ElMarsh for any delta, there exists an epsilon such that 0 < | x-a| < delta --> |f(x) - L | < epsilon
Property "P". You want to find a function f, and a point a, for which "f is P at a", but f is not continuous at a. Or conversely, f is continuous at a, but "f is not P at a".

Look at the floor function. It is not continuous at any integer (convince yourself). So let a be an integer and show that f is P at a.
Then you have a point for which floor(a) satisfies P at a, but floor(a) is not continuous at a; in other words, the definitions are not the same.

October 15th, 2009, 02:20 PM   #9
Senior Member

Joined: Apr 2009

Posts: 201
Thanks: 0

Re: counter example to a skewed deltaepsilon defns?

Quote:
Originally Posted by cknapp
Quote:
 Originally Posted by ElMarsh Hi, I think for the first part, if you have a discontinuous function that has very large intervals between each point or segment, you wouldn't be able to find an epsilon for a certain delta because the epsilon would have to span infinitely
The first problem with this is idea is that it doesn't work-- Those gaps need to be well-defined and fixed. For any given delta around a certain point, you will be able to find a "good" epsilon-- it will be very large, no doubt, but it will work.

That doesn't matter though: My suggestion was to look at a discontinuous function which satisfies the statement you gave. If the function is discontinuous, and doesn't satisfy the sentence, you get nothing, because you are trying to show that this definition doesn't line up with continuity. In other words, call
Quote:
 Originally Posted by ElMarsh for any delta, there exists an epsilon such that 0 < | x-a| < delta --> |f(x) - L | < epsilon
Property "P". You want to find a function f, and a point a, for which "f is P at a", but f is not continuous at a. Or conversely, f is continuous at a, but "f is not P at a".

Look at the floor function. It is not continuous at any integer (convince yourself). So let a be an integer and show that f is P at a.
Then you have a point for which floor(a) satisfies P at a, but floor(a) is not continuous at a; in other words, the definitions are not the same.
I kind of have an idea of what you're saying but I'm not so sure.. could you please clarify?

but if you have discontinuity, the limit doesn't exist anyway, you would have to define single sided limits, wouldn't you?

I'm thinking that the definition would be satisfied for the floor function, but since x->a takes on different limits, it can't be true (even though it is satisfied..)

thanks

 October 15th, 2009, 10:00 PM #10 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: counter example to a skewed deltaepsilon defns? I thought the problem was "Show that the statement "for any delta, there exists an epsilon such that 0 < | x-a| < delta --> |f(x) - L | < epsilon" is not equivalent to the definition of continuity." You have two options: 1) Show a function which is continuous, but does not satisfy the property above OR 2) Show a function which satisfies the property above, but is not continuous. I *think* functions that fit option 1 ore more degenerate than ones that fit option 2... I could be wrong though, and there might be a very nice function which is continuous but does not satisfy the statement. The floor function is one-sided continuous, but one-sided continuity is not continuity: it's one-sided continuity. So, the floor function is not continuous at any integer. If you can show that the floor function satisfies the above sentence, you've done number 2.

 Tags counter, defns, deltaepsilon, skewed

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post bboyinmartin Algebra 3 January 19th, 2013 05:48 PM Modus.Ponens Real Analysis 0 May 16th, 2012 02:57 PM aloria Real Analysis 2 January 25th, 2012 05:20 AM stangerzv Number Theory 4 January 10th, 2011 10:00 AM sixflags11804 Algebra 1 September 29th, 2008 11:26 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top