Math Is Fun Forum

Yep. When I first wrote the proof I just said "R\S = (-∞, 0.9) U (0.99, 0.999) U ..." without thinking. The problem is that saying this actually is the same as saying 1 ∈ S, since I said each set in the union is open! So I proved my assertion by unknowingly assuming its truth. Getting excited about having an interesting proof to a classic problem and writing it quickly like that is how you accidentally make false proofs . Your turn to post one, or you can pass it on to whoever else wants to if you have nothing.

Yeah, I noted that at the end of my post. I was thinking too hard at first . I'd consider a constant function to be a trivial answer though.

Where were you before San Francisco? Are you from another country and English is not your first language?

I guess I'll post one now. I came up with this for a certain thread which I'm sure you all know of... I won't mention its name here as it has fallen dormant and it might be best for it to stay that way. But anyway, a bit after I wrote it I thought about it again and realized it had an error. Now it's your turn to find it!

Theorem: Let S = {0.9, 0.99, 0.999, ...}. Then 1 ∈ S.

Proof: First, note that S is closed, since its complement R\S = (-∞, 0.9) U (0.99, 0.999) U ... is a union of open sets, and thus open. Also, 1 is an accumulation point (limit point) of S, since any open ball B(1, r) contains some point in S that is not equal to 1. But S is closed, and hence contains all of its accumulation points. Then 1 ∈ S.

That's a strange smiley there. But here's my thought on the situation. You've given us the illusion that {a[sub]n[/sub]} converges to a[sub]N + 1[/sub]. But we cannot conclude this. Why? Because for different ε, we may have different N. N is dependent on ε. Then a[sub]N + 1[/sub] may not always be the same number since N changes with a different choice for ε. But we know that if a limit of a sequence exists, it must be unique. Therefore we cannot conclude that {a[sub]n[/sub]} converges to a[sub]N + 1[/sub], since a[sub]N + 1[/sub] is not necessarily unique for all N.

Nice. What led you to pursue an MS in CS rather than mathematics?

Also in my opinion Proximity is the best game here... although I haven't really played any others since Proximity is addictive enough. Maybe someday I'll post my high score strategies.

I guess it's just another one of those "contradictory mathematical terminologies." Even if we revised it to be called a "open (n-1)-sphere" it isn't necessarily (n-1)-spherical region in a general n-dimensional metric space, as the metric can certainly give it a different shape. Oh well, the term has stuck with me.

1. Just take an open ball in R² centered at some point x ∈ (a, b). No matter what the radius of the ball, it will always contain points not in (a, b). Therefore there exists no 2-ball contained entirely in (a, b), so (a, b) is not open in R². This is because the open ball in R² is (in a standard sense) a circle, while the open ball in R is just an open interval.

2. He needed to "make it more complicated" because he cannot just say that since h is positive, q is an interior point. Imagine h getting smaller and smaller so that q is closer and closer to being a boundary point, so that d(p, q) approaches r. He does the other steps to show that q is never a boundary point. How can you reason this? Well he shows that any point s in an open ball of radius h centered at q is contained in E. Since q is arbitrary, we see that for any point in E this holds. This is precisely the definition of openness; that at any point of the set E you can center an open ball B of some suitable radius such that the ball is entirely contained in the set; so B is a subset of E. The proof just shows that the suitable radius is this h.

I think the biggest problem with getting good responses to your "lectures" is simply the audience here. There aren't so many people at the undergraduate level or above here, or they aren't interested in mathematics at that point. I for one enjoy your threads, even if I don't comment on them. It was sad to see your thread on Lie Groups end before you had even given a definition, as they are truly an interesting topic. Just know that even if I don't say anything I'm watching, I read everything posted in this section.

Have you ever even read Harry Potter?

It's just the group of integers modulo 6... The group operation is addition.

That will show that R is an infinite set, but it doesn't tell us whether it is countably or uncountably so. Q also has the property that between any two of its elements lies another number in Q. But Q is countable. So your method only shows that a set is infinite (by demonstrating that there are an endless amount of elements) but it does not address the concept of countability. It needs to show that there exists an indexing of the set using the natural numbers/one-to-one correspondence to N (or that there doesn't) to demonstrate countability (uncountability). You need to stick with the definitions as closely as possible; infinity can trip you up pretty easily if you try to work with it through intuition alone.