Find the indexed set $\bigcup_{\alpha \in \Bbb R} \{\alpha\} \times [0,1]$

Question

Find $A$ and $B$ where

$$A = \bigcup_{\alpha \in \Bbb R} \{\alpha\} \times [0,1]$$

and

$$B = \bigcap_{\alpha \in \Bbb R} \{\alpha\} \times [0,1]$$

Can anyone explain how to tackle this type of question? I have no clue to this... thank you.

Answer 1

$A$ comprises of joining together the many thin, non-overlapping vertical strips that extend from the $x$ axis $1$ unit upwards. So the end result is a band of width $1$ lying along the $x$ axis.

$B$ comprises of all the points that are common to all those thin vertical strips mentioned above. However, there aren't any common points, cause the strips are non-overlapping. So $B$ is empty.

Answer 2

The product $\{\alpha\} \times [0,1]$ is the set

$$V_\alpha = \{(a,b) : a \in \{\alpha\},\ b \in [0,1]\} = \{(\alpha,b) : b \in [0,1]\}.$$

Now $A = \bigcup_\alpha V_\alpha$ is the union over all of these. That is,

$$A = \bigcup_{\alpha \in \Bbb R} \{(\alpha,b) : b \in [0,1]\} = \{(\alpha,b) : \alpha \in \Bbb R,\ b \in [0,1]\} = \Bbb R \times [0,1].$$

In the other case,

$$B = \bigcap_{\alpha \in \Bbb R} V_\alpha = \bigcap_{\alpha \in \Bbb R} \{(\alpha,b) : b \in [0,1]\}.$$

However, if $\alpha_1 \ne \alpha_2$, then $V_{\alpha_1} \cap V_{\alpha_2} = \{(\alpha,b): \alpha = \alpha_1,\ \alpha = \alpha_2,\ b \in [0,1]\}$ is empty since $\alpha$ can't be both at once. Therefore $B = \emptyset$.