Yes, that is basically what I used. Anything with two horizontal asymptotes satisfies this condition. I was making it much harder than it needed to be.

Thanks. I may be asking you some more questions this weekend as I attempt the next problem set, so hopefully you'll be around

~BD

I already have some good procrastination scheduled then

Jackpot.

The answer I used for the previous one was x/(|x|+1) if you cared.

Also, do you mind me asking how you know all this math? Are you working in industry/research/teaching/in college?

~BD

Doing a PhD in math. My instinct is to use smooth functions where I can, but of course your solution is perfectly correct, and in some sense more intuitive.

Applied Math? or just math?

And yeah, my knowledge of functions is lacking somewhat. Many times I know what the function needs to look like but generating one is often times where I fall into some trouble. But, Im slowly increasing my function bank.

~BD

Pure math. I'm primarily interested in algebra, and my research is likely going to involve elliptic curves. I think the subfield would be called "arithmetic geometry."

Okay, Im struggling with this one:

Let f (x) = 0 if x is irrational and f (p/q) = 1/q if p and q are positive integers with no common factor.

a.) Show that f has a removable discontinuity at each positive rational number.

b.) Show that f is continuous at every positive irrational number.

~BD

Edit: Answered Thanks to Anilv

Here's part b. Fix an irrational x and epsilon > 0. We want to find a delta > 0 so that 1/q is less than epsilon whenever p/q is within delta of x. That is the definition of continuity at x. Equivalently, q > 1 / epsilon whenever | p/q - x | < delta. There is a finite number of q's less than 1 / epsilon, and each of these contributes a unique p/q that's closest to x for that q. Pick any delta > 0 that is smaller than this finite set of distances, and you are done.

For part a, since the rationals are dense in the reals, every rational number is arbitrarily close to the set of irrational numbers. Therefore, the function is not continuous at any p/q (since it goes between zero and q on an arbitrarily small open neighborhood). However, setting the function equal to zero on each rational would restore continuity by the previous argument, so the function has a removable discontinuity at each rational.

Subject: Honors English

I need to do a 3 page Research Paper about Neo Nazis or the Ku Klux Klan. I don't care wich one i do but i need a Topic Question thing about one of them. It can't be a simple question because i need to use three pages to write about.

I recall reading in Freakonomics that the KKK effectively came to an end in a pretty interesting way. Apparently, someone managed to get their secret sayings and handshakes and stuff into the parlance of kids of that time, which trivialized and satirized them. You could write a paper comparing the KKK and Neo-Nazism, and in particular explore whether this method could as effectively combat the latter. More generally, any sort of comparative essay would be pretty cool, if you could find the sources to back up your arguments.

Ah, okay. I wasnt sure if I could just say that for a and have the be sufficient. As for part b, that makes sense. I need to practice a lot more epsilon-delta proofs. Haha.

Okay, I have one more I've been struggling with as well:

(a) Prove: If f assumes only finitely many values, then f is continuous at a point x0 if and only if f is constant on some interval (x0 − δ, x0 + δ).

(b) Let M be a metric space and A ⊂ M . Define the characteristic function χA (x) for the set A as:

1, x ∈ A;
χA (x) = 0, x ∈ M \A.

Prove that χA is continuous at x0 ∈ M if and only if x0 ∈ int(A) ∪ int(M \A).


Im struggling to even figure out what is being asked in both parts.

~BD

If you aren't satisfied with my rigor at any point, you should be all means do the problem in more detail. Analysis isn't my strong point so I often gloss things I think are clear or simply too cumbersome to pursue further.

(a) If f is continuous, use the intermediate value theorem to show that f must be constant on some open interval. Otherwise, say f assumes a different value on that interval and construct an infinite number of points that f achieves between the two. Conversely, say f is constant on an open interval about x_0. Then by definition f is continuous at x_0.

(b) The nuance of this problem is that A might be a set that is neither open nor closed. We're supposed to prove that chi_A is continuous everywhere except the boundary of A (which might not all lie in A, if A is not closed already; recall the boundary of A is defined as the set of limit points minus the interior). We can use a generalization of part (a) to show that if x_0 lies in int(A) or int(M \ A), then chi_A is continuous at x_0 (you can find an open neighborhood around x_0 that is either all the way inside int(A) or all the way inside int(X \ A) ). On the other hand, if x_0 lies on the boundary of A, then by definition every open set containing x_0 has points in A and points in M \ A, so we use the definition of continuity to show that chi_A is not continuous there.

Note that if you turned in what I wrote and no more, you probably wouldn't get all the points on the assignment. However, I think everything you need is there.

Okay, Im still a bit confused on part a. Im not really sure where the Intermediate Value Theorem comes into play.. I guess Im just unsure of the question as a whole.

As for part b.) That makes a lot of sense now. Would this be a correct thought process for part b?

We could look at the right and left approaching limits of x0 in Chi_A. x0 would be continuous if (and only if?) the two limits were equal to one another. So looking at

Lim (x-> x0+) of Chi_A must = Lim (x-> x0-) of Chi_A. Where x -> x+ is the limit from the right and x -> x0- is from the left. Granted I dont know if you can take a limit in a metric space but thats how I was visualizing it.

Also, I plan on reading your answers and then walking away from my computer and formulating my own, as to make sure I dont copy . Copying would be a waste of yours and my time as I wouldnt be learning anything.

~BD

Good for you. As to your questions:

(a) I've thought about a bit more and decided that the intermediate value theorem won't work, so sorry about that. The problem with it is that you would need f to be continuous on an interval around x_0, which is not a priori true. The following works, however.

The only way for there to not exist some open interval of x_0 on which f is constant is if there is a sequence {y_i} converging to x_0 with f(y_i) ≠ f(x_0) for all i. That is, we need f to be non-constant arbitrarily close to x_0. Supposing there exists such a sequence, we will solve the problem by deducing that f is not continuous at x_0 (proof by contrapositive, if you like). If f were continuous at x_0, then the image of the sequence { f(y_i) } would converge to f(x_0) since {y_i} converges to x_0. You can prove this result separately if you don't know it. We could then assume that { f(y_i) } is a strictly monotone sequence; if it weren't, we could just throw away the y_i's that keep it from being monotone. But this sequence has infinitely many distinct values, which is impossible by hypothesis. Therefore, f can't be continuous at x_0.

Again, the other direction is easy since constant implies continuous.

This proof is a bit more round-about than I would like, but it does get the job done. Hopefully you can pare it down to the essentials (getting rid of the bit that argues by contradiction would be a start).

(b) You're thinking along the right lines but you'll have to raise the level of abstraction a bit for this problem. Metric spaces are much more general than the real line; for example, n-dimensional Euclidean space (R^n) is just one example of a metric space, and even in this case the notion of left- and right limit is gone. To solve this problem you'll need a better definition of continuity, such as the following.

f is continuous at x_0 if and only if:
for all epsilon > 0 and all y in M, there exists delta > 0 such that || x - y || < delta implies | f(x) - f(y) | < epsilon.

I use the || || symbols to remind you that x and y can be vectors (or even more exotic objects), and what really matters is that we have a metric that tells us the distance between them. Note that the set of all y such that ||x - y|| < delta is also called the open delta-ball around x, for reasons I hope you can see.

You should be able to generalize the result in (a) to this metric space context without much trouble. All you need to use is that int(A) is an open set so every point in it is enclosed in an open neighborhood lying entirely in int(A). In particular, for all x in int(A), there is some delta > 0 such that the delta-ball about x lies in int(A). Since chi_A is constant on A, it is definitely continuous at every point in int(A) for this reason. The same argument holds for int(M \ A) since this is also an open set and chi_A is constant here too. This is the easier direction.

If you accept the definition I gave for the boundary of A (denoted bdy(A) ), then you should agree that if x lies in bdy(A), every open neighborhood of x contains points in A and points in M \ A. If you're not sold, consider that bdy(A) = bdy(M \ A) so x is a limit point of both A and M \ A. In particular, this is true of every open delta-ball around x. You should conclude from the definition of continuity that f is not continuous at x.