# Visualizing $\cap_{i = 1}^\infty A_i = (\cup_{i = 1}^\infty A_i^c)^c$

As the question title suggests, how do I visualize$$\cap_{i = 1}^\infty A_i = (\cup_{i = 1}^\infty A_i^c)^c?$$Let's start with the right-hand side. So I have a bunch of circles representing the $A_i$'s, right? Then I take the complements of each one, so for each $A_i$, the corresponding $A_i^c$ is the entire ambient space with the $A_i$ removed. I have trouble visualizing that taking the union of all those $A_i^c$'s then taking the complement of that union is our desired intersection $\cap_{i = 1}^\infty A_i$. Could anybody help me visualize this?

• how do you even "visualize" $\cap_{i=1}^{\infty} A_i$? :P
– user290300
Commented Aug 23, 2016 at 14:15
• I think it's easier to consider the equivalent $$\left( \bigcap_{i=1}^\infty A_i \right)^c = \bigcup_{i=1}^\infty A_i^c$$ Commented Aug 23, 2016 at 14:20
• Also, I would tend not to "visualize" this at all; to me, this comes down to logical definitions. In particular, I would prove it as is done here. Commented Aug 23, 2016 at 14:23
• In my opinion, don't visualize an infinite set. Reduce this to just two sets and think about the Venn Diagram. Commented Aug 23, 2016 at 14:30

A graphical example of Robert Israel's explanation.

Some sets $A_1$, $A_2$, $A_3$, $A_4$ and their intersection $\cap_i^\infty A_i$.

Now we calculate their complementary (only the shaded parts).

Finally we join them and calculate the complementary of the resulting set.

We see that $\cap_i^\infty A_i = \big(\cup_i^\infty A_i^\complement\big)^\complement$

• This is an amazing answer -- what software did you use to make these pictures/where did you find them? Commented Aug 23, 2016 at 15:10
• Thank you. I've done it with CorelDRAW (it is propietary software). However, I suppose that it wouldn't be that difficult to do it with a free one such as Inkscape, for example.
– JnxF
Commented Aug 23, 2016 at 15:13

It's easier to visualize $4$ than $\infty$, but the principle is the same.

Draw $4$ circles (and make sure the intersection of all $4$ is nonempty). Note that this intersection is the only region that is not outside any of the circles.

It’s easier, I think, to let $B_i=A_i^c$, so that your identity becomes $\bigcap_iB_i^c=\left(\bigcup_iB_i\right)^c$. Now draw your picture, with circles for the sets $B_i$. The set $\bigcap_iB_i^c$ consists of those points that are not in any of the circles, and so does the set $\left(\bigcup_iB_i\right)^c$.