Math Help.

[Jack Squat]

I'm having a problem with this. I'm supposed to prove the following:

For all ε > 0 there is an a > 0 such that a < ε

The way I'm seeing it, there can't be a value for a that is less than every ε because they both need to be greater than zero. I'm sure I'm missing something obvious, anyone care to point it out?

[afroakuma]

Given that ε > 0, ε/10 is always smaller than ε.

a = ε/10

There's no flaw here at all. Even if ε is 0.000001, a could just be 0.0000001 and would be both greater than zero and smaller than ε.

[Tirian]

You are correct that there cannot be a positive a that is less than every positive ε, but that is not what you are being asked to prove. The problem is, given a positive number ε, find a positive number that is less than it. This is easy, all you have to do is prove 0 < ε/2 < ε.

[Jack Squat]

ah...I just read it wrong. I was reading it as saying there needs to be a value a that's less than all values of ε. Thanks for clearing that up.

I'm really going to need to turn off the part of my brain that "interprets" reading for this class.

I'll leave this open to others to post their questions then, to try and avoid clutter.

[afroakuma]

There is a pseudo-solution to that, IIRC, involving limits, but it ain't fun.

[Tirian]

I'm really going to need to turn off the part of my brain that "interprets" reading for this class.

No such luck. You took the red pill, and now you're in the world of delta-epsilon proofs. They'll be with you all the way through to advanced calculus. They're not so hard to read once you get more experience with them, though.

[afroakuma]

I seem to recall being given a class-wide bye on delta-epsilons.

[Jorkens]

No such luck. You took the red pill, and now you're in the world of delta-epsilon proofs. They'll be with you all the way through to advanced calculus. They're not so hard to read once you get more experience with them, though.
Yeah, it's sort of alright once you realise that epsilon / delta proofs are essentially a bunch of very very obvious statements translated into a rather nonobvious language.

Or to be more accurate, that it's checking that the very very obvious statements work as well in your nonobvious language as they intuitively ought to.

[Tirian]

I seem to recall being given a class-wide bye on delta-epsilons.

It depends on the class. The proofs are at the low level of basic limit proofs and the concept of continuity, which come up in pre-calc and all three semesters of calculus. I suppose when I learned the Chain Rule the professor made it clear that we wouldn't be expected to recreate the proof on a test, so to a certain degree you can just assume that there IS a d-e proof and leave it at that. But then I had to prove it in my Differential Geometry course for abstract manifolds, which was no fun at all.

Once the notion is exposed to you, you're probably going to be faced with a bunch of problems like graphing (x² - 2x + 1)/(x-1) and proving that it is continuous at 1 even though the value isn't defined there, and you should know the concept enough to be able to look at the overview of a d-e proof in a textbook and understand why it is trying to make the case that it is.

[afroakuma]

I've long since forgotten most of calculus and precalculus, more's the pity. Bad uni prof going over it again will do that to you. :smallfrown:

[mathemagician]

When I teach this stuff, I've found it occasionally useful to think of these things as a 2-player-game.

This was your problem:

For all ε > 0 there is an a > 0 such that a < ε

Step through each bit of the problem. Player 1 will be "the bad guys" and you will be player 2. So the game might go like this:

Player 1: Here's an ε, let's say ε=10. I bet you can't find a > 0 and a < ε.
Player 2: I'll elect a = 2.
Player 1: Fine! How about ε=1.3. I bet you can't find such an a now!
Player 2: How about a = 1.
Player 1: Well how about ....

Then the real problem becomes, can you prescribe a strategy such that no matter what ε Player 1 gives you, you can automatically write down the a? In this case, a = ε/10 or a = ε/2 are correct "strategies." It is of course important that you realize that it has to work for every ε, because just finding values for a for different examples of ε is not enough.

----

Probably the best advice I can give for proofs, is to write down a few examples of the statement in question to "see why it works," and then try to prove it. If you can't make progress, the next step I always take is to try to find a counter example; I find that in my desperate attempts to prove something is not true, the things that go wrong are often the key steps steps in the proof.

----

And one last comment, if I may, since it relates to how you misinterpreted the question. It has to do with the order of the "for all" and "there exists." When "for all ε" is the first line of the statement, it means that while ε can be anything, BUT, for the rest of the statement it is "fixed." It is what we mathematicians often call "variable but fixed." You fix the ε, and then proceed with the rest of the problem. Next up is the "there exists." It says "for that particular ε" there is some a with such and such property.

What if the problem was backwards?

There is an a > 0 such that for all ε > 0, that a < ε .

The goal in this problem would be exactly what you were trying previously. You try to find an a > 0, such that no matter what ε is, you have a < ε . You see, that when "for all" follows "there exists" your goal becomes to find just one magical a, but it has to work for every ε imaginable. A lot tougher, unless you change it to a >= 0! ;)

So the order is very very important!

I hope this gives some insight. The best way to get these things straight is to write down examples of the statement first, and do lots and lots of problems!

Sorry if you didn't need all that help, or if it's a little preachy. Since being on a research grant, I haven't been able to teach, and well, you asked about my favorite class to teach... :P

[Jack Squat]

Alright, I've got another one that I could use you guys on.

A, B, C, & D are sets. Prove (AuB)x(CuD) = (AxC)u(BxD)

What I have so far is on proving c is:
Let (x,y) = (AuB)X(CuD) = (AXC)u(BXD)
By definition of X, x ε AuB and y ε CuD

I know I can go through and use cases of xεA, xεB, yεC, and yεD, but I was wondering if I could instead just skip and go:

By definition of u, xεA or xεB and yεC or yεD
Hence, xεA and yεC or xεB and yεD
By definition of X, (x,y) ε AXC or BXD
By definition of u, (AXC) u (BXD)

Then all I'd have to show for the other way is to reverse the steps.

Thanks for any help.

[Tirian]

I spent a semester as a paper-grader for Disco, and if you turned in something that brief and formalistic to me, I would be dubious as to whether you understood the logic. Use cases is probably a better way to demonstrate your skills at least until later in the semester.

To give a rather stunning illustration of that point, your brief proof didn't help you to notice that the statement is untrue. :smalltongue: If A = D = Ø, then in general (A˅B) × (C˅D) = B×C ≠ Ø = Ø˅Ø = (A×C)˅(B×D). It is true if you replace the unions with intersections, or you could show that (A×C)˅(B×D) is a subset of (A˅B)×(C˅D).

(Argh, can someone tell me why there is a Unicode symbol for intersections but not unions? Do you think that u looks like a union symbol?? :smallconfused:)

[Jack Squat]

I spent a semester as a paper-grader for Disco, and if you turned in something that brief and formalistic to me, I would be dubious as to whether you understood the logic. Use cases is probably a better way to demonstrate your skills at least until later in the semester.

To give a rather stunning illustration of that point, your brief proof didn't help you to notice that the statement is untrue. :smalltongue: If A = D = Ø, then in general (A˅B) × (C˅D) = B×C ≠ Ø = Ø˅Ø = (A×C)˅(B×D). It is true if you replace the unions with intersections, or you could show that (A×C)˅(B×D) is a subset of (A˅B)×(C˅D).

(Argh, can someone tell me why there is a Unicode symbol for intersections but not unions? Do you think that u looks like a union symbol?? :smallconfused:)

D'oh! :smallredface: I've been going over boolean logic as well this semester so my half-asleep self figured "if the example he gave us using intersections (ands) is true, the unions (ors) are true as well" Didn't even think to run examples. I suppose I learned a valuable lesson - don't get two and a half hours of sleep before you have to do some thinking :smalltongue:

My guess on why there's not a symbol for unions is because the intersection symbol is an upside down capital U. (as a note of comparison U ∩), so they figure you'll just use that. I typed it with lowercase to make it easier to read, since the sets were capital.

I'm kinda baffled why they don't have subsets on the character map...I suppose they guess you'll just use a c and then underline it if it's not proper.


Thanks again. I'll be back before too long

[Tirian]

Oh, I guess so. I've got Times New Roman as my default font for charmap, so my U has serifs on it and looked very unacceptable. I am so badly spoiled by TeX.

Yes, the lesson for the day is that intersections almost always behave much better than unions, and that turns out to have many applications across mathematics. For instance, a convex set in geometry is a plane figure such that for any two points in the figure, the line segment connecting those two points is also entirely inside the figure. So, using that definition you can show that the intersection of a (non-empty, possibly infinite) collection of convex sets is also convex. Because of that lemma that convexity is "intersection-stable", you can easily prove the theorem that any plane figure has a unique smallest convex superset, called the convex hull. It's easy to prove because the convex hull of a figure is simply the intersection of all convex supersets of the figure; obviously that is both convex and there can't be a convex superset that is smaller than the intersection of all of them.