Recent content by fopc

I said nothing of the sort. Nobody said (or even suggested) "they both must have dimension n-1". I suggest you try reading the post again--carefully this time. If he's given that one hyperplane has dimension p, then clearly it resides in a p+1 dimensional vector space. That would...

In the given context, I'd say a hyperplane of vector space V, where dimV = n, is an (n-1)-dimensional linear manifold. From here you should be able to arrive at the answer given in the back of the book.

If I understand you correctly, we now have injections, and you ask for a bijection? Schroder-Bernstein. The classical proof of this theorem is a constructive-existence proof (no AC). But contrary to what was suggested in another post, the construction of the bijection is non-trivial (my...

If by "the minimal element of f inverse of x" you mean "the min of the pre-image of the unit set {x} under f", then OK. Of course, I don't what else you could have possibly meant. "1 is not possible without AC." Fine. Can't say I'd disagree with you. With a slight generalization it can be...

Here are two problems. One will fit on your list, the other won't. Preliminaries: R denotes set of real numbers N denotes set of natural numbers 1) f:R->N arbitrary surjection. Show there exists an injection g:N->R s.t. f(g(b)) = b for every b in N. 2) f:R->R arbitrary surjection...

Here's an example of how a "quotient" map might be constructed. Let f:A->B (f an arbitrary map, A and B arbitrary sets). The map f induces a natural partition on its domain, A. The equivalence relation associated with this partition is defined by, a~b iff f(a)=f(b). Take ~ as your...

Look up "subadditive".

I'll assume you're starting from: f:A -> B and A_0 is a subset of A. The inclusion relation you've written holds regardless of whether f is injective or not. However, if f is injective, then the relation can be written as an equality. Proof is nothing more than working the definitions, as...

Maybe you've already proved it. If not, here's a suggestion. Put the statement into a form that'll shed some more light on the different options you have for proving it. The first thing I'd do is get rid of the "unless"; then you'll have a positive statement. It'll be an implication...

I suspect you didn't read the EDIT. It reads, f(x) = x for all x in [0,1]-A. Not mathematically rigorous? Well, I defined an f, and made a claim that f is a bijection; nothing more. I then suggested the need to demonstrate this by showing f is injective and surjective. There's hidden...

Here's a plausable proposition. Every infinite set includes a denumerbable subset. Proof is straight forward with AC. What about without AC? Here's a couple of definitions. A set S is finite if there is a bijection f:{0,1,...,n-1} -> S. A set is infinite if it's not finite. A set...

Apparently no details are given by the author (even in the exercise statement). So you'll have to make some assumptions. It's a hypothetical formula, so I think it's safe to say it's universally quantified. No problem. But, the problem domain is another matter. I suspect the author meant...

You might look at a proof this way. Let's let |S| = n and S' = S U {a} (where a is not in S). Clearly, |S'| = n+1. Let P(S) and P(S') denote their powersets. Let A = {B U {a} | B is in P(S)}. In the induction step, the induction hypothesis says |P(S)| = 2^n. So if you can show the...

Loomis and Sternberg, chapter 3 section 5 and following sections. You'll have to do some preliminary reading in order to get to this point. Their text is freely available. Even if you don't do more than look at it now, it'll serve you well in November.

I don't know what preliminary steps the author took in an attempt to ensure the non-replacement claim. You haven't provided enough information.