Recent content by mikemhz

Sooo.. X = \left\{1, 2\right\} Y = \left\{A, B\right\} Z = \left\{I, J\right\} X × (Y × Z) = \left\{(1, (A, I)), (1, (A, J)), (1, (B, I)), (1, (B, J)), (2, (A, I)), (2, (A, J)), (2, (B, I)), (2, (B, J))\right\} Take any element from X × (Y × Z): e.g. (2, (A, I)) Show that there is...

Would this show that it is injective? x, a \in X y, b \in Y z, c \in Z \varphi((x,y),z) = \varphi((a,b),c) \varphi : (x,(y,z)) = (a,(b,c)) x=a, y=b, z=c End of proof. How about surjective? To be honest, I'm not 100% what I need to prove in each case. I know that injective is...

Show that for any sets X, Y , Z, the canonical function: \varphi : (X × Y) × Z \rightarrow X × (Y × Z) (\varphi((x, y), z) = (x,(y, z))) is a bijection. Solution. We can do this by showing that \varphi is injective and surjective.. I can do this by showing \varphi has an...

So I just looked through the lecture slides and the "+" symbol means the sum or disjoint union. X+Y = { (x,0) | x∈X} ∪ {(y,1) | y∈Y} I think this is an important point to clarify because it crops up again later in the module. Still why 0 and 1? EDIT: OK. In a moment of clarity I've...

So the full question is phrased like so: Suppose X; Y and Z are sets. Does X × (Y + Z) = X × Y + X × Z? If X, Y and Z are finite, what can we say about the cardinalities of X × (Y + Z) and X × Y + X × Z? I'm not looking for the solution. It's sitting here in front of me. I just don't...

Set Theory "+" symbol 1. X, Y and Z are sets. Does X × (Y + Z) = X × Y + X × Z? The solution starts like so: X × (Y + Z) = {(x,(y,0)) | x \in X, y \in Y}\cup{(x,(z,1)) | x \in X, z \in Z} I don't understand how the "+" symbol works. Why does it equate to this (x,(y,0)) (x,(z,1))...