The union of two subsets of $\mathbb R^3$.

Question

I have the following subsets of $\mathbb R^3$: $$A=\left\{(x,y,z): \ z_1\leq z\leq z_2\, , \ x<\bar{x}\, , \ y_1\leq y\leq y_2\right\}$$ $$B=\left\{(x,y,z): \ z_1\leq z\leq z_2\, , \ x>\bar{x}\, , \ y_3\leq y\leq y_2\right\}$$ where and $\bar x$, $y_1$, $y_2$, $y_3$, $z_1$, $z_2$ are fixed numbers and $y_1\leq y_3$.

What is the union of the two sets? I thought: $$A\cup B=\{z_1\leq z\leq z_2\, , \ x\neq\bar{x}\, , \ y_1\leq y\leq y_2\}\, .$$

Answer 1

It might help to write the sets as

$$A = [z_1, z_2] \times (-\infty, \overline{x}) \times [y_1, y_2]$$ $$B = [z_1, z_2] \times (\overline{x}, +\infty) \times [y_3, y_2]$$

so $$A \cup B = [z_1, z_2] \times \Bigg( \Big((-\infty, \overline{x}) \times [y_1, y_2]\Big) \cup \Big((\overline{x}, +\infty) \times [y_3, y_2]\Big)\Bigg)$$

It cannot really be simplified any further.

Answer 2

As pointed out by Andreas Blass in the comment, your approach isn't correct, indeed it should be

$$A\cup B=\{(x,y,z):z_1\leq z\leq z_2\, , \ x<\bar{x}\quad \text{for}\quad y_1\leq y\leq y_3,\ x\neq\bar{x}\quad \text{for}\quad y_3\leq y\leq y_2,\}$$

Answer 3

I thought it might be helpful to have a graphical illustration of what's being joined here:

Both blocks go off infinitely far to the left and right, since there is no lower $x$ bound for the first subset, and no upper $x$ bound for the second subset.

Note the small gap between the blocks at $x = \overline{x}$; the union does not contain any points for $x = \overline{x}$.