Union Set and Correspondence In $R_{+}^{2}$ I have these sets: $q = [0,1]$ and $Z=\{1,2\}$, Let $\mathbb{Y}$ be the set combination $(q,z)$ .
Let Y and Y' elements of $\mathbb{Y}$.  $Y = [0,1] \times \{1\} \cup \{(0,2)\}$ and $Y' = [0,1]  \times \{2\} \cup \{(0,1)\}$
My questions:
1) Y is a vertical line on 1? And Y' is a vertical line on 2?
2) It is clear that there is no relation such $\subseteq$ or $\supseteq$ between $Y'$ and $Y$. Right?
3) What Is the joint set: $Y \cup Y'$ and the intersection set: $Y \cap Y'$?
What I did:
1) I believe it is right that there is no relation $\cup$ or $\cap$ between $Y'$ and $Y$.
2) This what I did: 
$Y \cup Y' = [0,1] \cup \{1,2\}$ It would be a vertical line on 1 and a vertical lie on 2?
 A: As it stands, there are a few things that don't quite make sense about your post, but I think I can tell what you're getting at. It seems that you mean the following:


*

*$Q$ is the real interval from $0$ to $1,$ inclusive.

*$Z$ is the set whose elements are just $1$ and $2.$

*$\Bbb Y=Q\times Z.$
Now, it looks like we have that


*

*$Y$ is the set comprised of the ordered pair $\langle 0,2\rangle$ together with the elements of $Q\times\{1\}.$

*$Y'$ is the set comprised of the ordered pair $\langle 0,1\rangle$ together with the elements of $Q\times\{2\}.$
If I'm correct about what you mean by $\Bbb Y,$ this would make $Y$ and $Y'$ subsets of $\Bbb Y,$ rather than elements of $\Bbb Y.$ I can't think of an interpretation of $\Bbb Y$ that would make $Y$ and $Y'$ elements of it, but there may be something I'm missing.
So, we could think of $Y$ as a vertical segment on $1,$ together with a single point off that segment. We can think of $Y'$ similarly. It's quite correct that $Y\not\subseteq Y'$ and $Y\not\supseteq Y'.$ $Y\cup Y'$ would consist of a vertical segment on both $1$ and $2.$ Moreover, if I'm correct about what you mean by $\Bbb Y,$ we in fact have that $Y\cup Y'=\Bbb Y.$
