Recent content by mathboy

Wow! Excellent guys! Does anyone dare give an explicit partition of the power set P(R) of the reals into |P(R)| disjoint subsets each with cardinality |P(R)| ?

Ah, yes. That does it! Now since |R| = |R|x|R|, then that means that R (reals) can be partitioned into |R| many disjoint subsets each with cardinality |R|. What is an explicit such partition of R? R can be partitioned into ..., [-1,0), [0,1), [1,2), [2,3), ... , each subset with...

I already know that for any two cardinal numbers a,b, that ab=max{a,b}. I am wondering if the partition is always possible for any infinite set A.

Yes, that was my original question (but I've generalized it now). How long is the proof in your textbook for this? What's the sketch of the proof? Does it depend on axiom of choice? Almost gives me what I seek. My generalized conjecture is: If A is infinite, then A can be written as...

I'm not trying to prove that |B|+|C| = |A|. That follows immediately from |A| + |A| = |A|. I'm trying to prove that B and C exist such that B and C partition A and each have cardinality |A|. But I think I've already proven it now, and am wondering about the infinite partitioning case.

Partition A into B U C, such that |B|=|C|=|A|. Actually, we should be able to generalize this to k partitions of A such that each partition has cardinatility |A|, and k is any (infinite) number less than |A|, right?

Here's my proof attempt without using ordinals. Let A={a_i| i in I}. Well-order I. Suppose all elements a_i can be placed alternatingly in bin B and bin C for i < j. Then a_j can be placed in B if the previous was placed in C (and vice versa). So by transfinite induction, this can be done...

So ordinals must be used to prove this?

and B and C must be disjoint. e.g. A=Integers: B= even integers, C=odd integers. Done. A=Reals: B=(-infinity,0), C=[0, infinity). Done. A=P(R): B=? C=? The problem begins with cardinality greater than c. How about this: Well-order A. Put the first element in the left bin, the...

So if A is an infinite set, we know that |A|+|A|=|A|. But are we allowed to go backwards, i.e. divide A into two disjoint subsets B and C such that A = B U C, and |B|=|C|=|A|. For the integers and for the reals, this is clear, e.g. R = (-infinity,0) U [0, infinity), each with cardinality c...

That's my opinion only.

How much does high school contest skill improve your grades in university math courses? Currently I find that all the contest tricks that I've learned play almost no role in my university courses, and studying my textbooks is all that matters now. Any clever thinking I had for those contests...

My question even stumped topologist Henno Bradsma. He said: "I found a result that a metric space has a minimal subbase (proved by van Emde Boas). So probably not all spaces have them... All finite spaces have a unique minimal base. This is all U_x, where U_x = /\{U: U open and x in U}...

How about this weaker assertion: Every topology T contains a minimal basis B for T (in the sense that any proper subset of B is not a basis for T). This must be true, right? And the same for subbases? But Zorn's Lemma still doens't work.

I can't seem to find this result in any of my textbooks. Given any basis B for a topology T on X, is there a minimal subset M of B that also is a basis for T (in the sense that any proper subset of M is not a basis for T)? If so, is Zorn's Lemma needed to prove this? Is the same true of...