Exponentiation of Sets Proof

So I am currently attempting to dive into Axiomatic Set Theory but before that I am trying to get a good solid base of Naive Set Theory.

I am stuck on how to prove the following three Theorems:

If XY∪Z∼XY X^{Y \cup Z} \sim X^Y x XZX^Z

If (X1X_1 x X2X_2)Y^Y ∼XY1\sim X_1^Y x XY2X_2^Y

If XYxZ∼(XY)ZX^{Y x Z} \sim (X^Y)^Z

Where ∼\sim means that there exists a bijection.

My attempt at the first proof:

If F∈XY∪Z F \in X^{Y \cup Z} then F:Y∪Z→X F: {Y \cup Z} \to X

where f1:Y→X f_1 : Y \to X and f2:Z→Xf_2 : Z \to X

and G∈ G \in XYX^Y x XZX^Z
where g1:Y→X g_1: Y \to X and g2:Z→Xg_2: Z \to X
then G(g1,g2)=hG(g_1 , g_2) = h

I am currently unsure as to what g is.

My attempt at the second proof:

If f∈ f \in (X1X_1 x X2X_2)Y^Y then f:Y→ f: Y \to (X1X_1 x X2X_2)
and g∈ g \in XY1X_1^Y x XY2X_2^Y then g: g: ?

My attempt at the third proof:

Let ϕ:XYxZ→(XY)Z\phi: X^{Y x Z} \to (X^Y)^Z

then ϕ(F)=G,G|y(z)=H,H(y)=F(y,z)\phi(F)=G, G\vert_y (z) = H, H(y)= F(y,z).

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Are YY and ZZ disjoint?
– Antioquia3943
Oct 21 at 3:07

  

 

Yes, and if it were the case that they were not, from my understanding, you can create a bijections that will make said sets dijoint.
– El Spiffy
Oct 21 at 3:09

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2 Answers
2

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Take F:XY×XZ→XY∪ZF: X^Y \times X^Z \rightarrow X^{Y \cup Z} such that if f:Y→Xf: Y \rightarrow X and g:Z→Xg: Z \rightarrow X then F(f,g)=hF(f,g)=h where h:Y∪Z→Xh: Y \cup Z \rightarrow X is such that h|Y=f\left.h\right|_Y = f and h|Z=g\left.h\right|_Z = g.

Take F2:XY1×XY2→(X1×X2)YF_2: X_1^Y \times X_2^Y \rightarrow (X_1 \times X_2)^Y such that if f:Y→X1f: Y \rightarrow X_1 and g:Y→X1g: Y \rightarrow X_1 then F2(f,g)=hF_2(f,g)=h where h:Y→X1×X2h: Y \rightarrow X_1 \times X_2 is such that h(y)=(f(y),g(y))h(y) = (f(y), g(y)).

  

 

Thank you. Thinking about it a little more I started editing it and then saw this response.
– El Spiffy
Oct 21 at 3:21

  

 

The third one is harder, can’t think in a bijection right now.
– Antioquia3943
Oct 21 at 3:23

  

 

@Antioquia3943 BIG ERROR. I meant the cartesian product of the sets, not the union. The union would never work.
– El Spiffy
Oct 21 at 3:39

  

 

Actually i think it’s correct with union. If they’re disjoint. Disjoint union if they’re not.
– ziggurism
Oct 21 at 3:49

  

 

@ziggurism I am speaking in reference to the third Theorem. I initially had X^{Y\cupZX^{Y\cupZ but I meant cartesian product of the sets.
– El Spiffy
Oct 21 at 3:50

Well if ff is defined on Y∪ZY\cup Z then ff can take as input elements of both YY and ZZ. To construct an element of XY×XZX^Y\times X^Z you need two maps, one which acts on elements of YY and another which acts on elements of ZZ.

For the second attempt, again you need to create a pair of maps. One from YY to X1X_1 and another from YY to X2X_2. Remember that ff will give you pairs of elements in X1×X2X_1\times X_2.

  

 

Thank you, this really helped get the ball rolling, I just edited my question.
– El Spiffy
Oct 21 at 3:22

  

 

You have to define the map f1f_1. What value should f1(y)f_1(y) take for y∈Yy\in Y? Remember to use your given data FF.
– ziggurism
Oct 21 at 3:49