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))...