Recent content by andytoh

Any suggestions on how to cope with foreign languages in journal papers? Say you want to read a proof of a certain theorem. But the ONLY published proof of that theorem is in Russian, and it has never been translated in English. Other than learning Russian, any suggestions on how to learn the...

Interesting. I suppose for every 10 mathematicians trying to prove a very believable conjecture, there should be 1 trying to find a counterexample. Without these counterexample experts, we would be forever trying to prove something that is not true!

That example qualifies I suppose. It was the #1 unsolved problem back 1900. And it Cantor did try to prove it till his death. Many topologists tried to prove that a countably infinite product of R is normal in the box topology. It's been quite a few decades of effort by many mathematicans...

When I say almost true, I mean that many mathematicians believed the conjecture was true, and for the longest time they all struggled really hard to prove it and the (long) proofs almost made it through, except for one line that wasn't quite right. They all tried to fix that one line, and...

f:Nx[0,1) -> [1,infinity) given by f(n,x)=n+x is your desired bijection.

http://www.math.ku.dk/~moller/e02/3gt/opg/S23.pdf Look for #12.

How about m U { (a, x), x in X such that (b, x) in m } U { (x, b), x in X such that (x, a) in m }.

But if we define a<b, and we have c<a from our maximal partial ordering (for example) but no relation between c and b in our maximal partial ordering, then transitivity also fails because we don't have c<b when we are supposed to. So we can define c<b, but I've checked that this runs into...

Thanks. But suppose we define a<b, and we have c<a but there is no relation between c and b at all? (remember that so far we only know that the maximal element is a partially ordered set only), or if we have b<d but there is no relation between a and d? Same problem if we define a>b.

If r is partial ordering on X, prove that r is contained in a total ordering on X. Hint: Consider the collection of all partial orderings containing r. Use Zorn's Lemma. I've already proven using Zorn's Lemma that there exists a maximal partial ordering m containing r. But I can't seem to...

Your solution is not correct because you didn't separate Y properly (read the precise definition of separation). The correct solution is more complicated than that. The connectedness of X is needed.

Are you sure you separated Y correctly? You will need to use closures in your proof. Make use of the fact that if two sets form a separation, then each set does not meet the closure of the other, i.e. ClA is in X-B, ClB is in X-A, ClC is in X-D, ClD is in X-D. Suppose YUA is not...

Suppose YUA is not connected, with separation CUD. Since Y is connected, then Y must be a subset of C or D, say Y is a subset of C. Now show that X is not connected. In the end, you will find that BUC and D are closed subspaces, and thus X = (BUC)UD is a separation of X.

f:R->{0,1} f(x)= (chi_S)(ln(x+1)) if x is a non-negative integer (chi_S)(lnx) if x>0 and not and integer, (chi_T)(ln(-x)) if x<0. Yuck. Can someone think of a better one?

Bijection F:P(R) x P(R) -> P(R): Start with two subsets S and T of R. Get the characteristic functions (chi_S, chi_T). Get the characteristic function f:R -> {0,1} defined by f(x)= (chi_S)(lnx) if x>0, (chi_T)(ln(-x)) if x<0. (problems with x=0, so someone find a better bijection from...