Why is it obvious that if the intersection of all sets is empty then its complements cover the whole space?

Question

I was watching Harvey Mudd's real analysis video talking about the finite intersection property. The first step of the proof relies using contradiction and thus starts by saying that the intersection of all the sets is empty:

$$ \cap_{\alpha } K_{\alpha} = \emptyset $$

Then the professor says the following (perhaps paraphrased a little bit):

If a point is in the intersection then it means that point is not in any of the complements. So if there is no point in any intersection then some point must be in some complement.

The first part of the statement seems obvious to me, if one is in all the sets then thats the same as saying that point must be in every set simultaneously. If one is in all the sets simultaneously then you can't be in any one of the complements (since otherwise you'd not be in that set and thus can't be in all of them simultaneously). Basically this relies that being in a set means you can't be in the complement of that set and being in all the sets means you can't be in any one of the complements. Basically:

$$ x \in \cap_{\alpha} K_{\alpha} \implies x \not \in\cup_{\alpha} K^c_{\alpha} $$

or

$$ \cap_{\alpha} K_{\alpha} = ( \cup_{\alpha} K^c_{\alpha} )^c $$

However, the argument proceeds to then say:

So if there is no point in any intersection then some point must be in some complement.

which is the part I am struggling to understand. I wish to understand it sort of in plain english. What they call sometimes "intuitively" because that second part of the conclusion is not something that would have occurred to me and it seems something that should have flowed naturally from a train of thoughts rather than some symbol crunching game.

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The reason I say not from a "symbol crunching game" is because its quite trivial to just start applying negations and complements until you force the answer out. e.g.

$ \cap_{\alpha } K_{\alpha} = \emptyset $

using DeMorgan's $ (\cap_{\alpha } K_{\alpha})^c = \cup K^c_{\alpha}$ and complementing it we get:

$ \cap_{\alpha } K_{\alpha} = (\cup K^c_{\alpha})^c = \emptyset$

then taking the complement of everything we get:

$\cup K^c_{\alpha} = \emptyset^c = X$

i.e. the complement of nothing is everything and then we get the desired answer. The reason I don't like this reasoning is because it just feels like symbol crunching rather the way professor Su explained it seems there is a naturally flowing way to reach that conclusion from logic or just reasoning without requiring long winding manipulation of rules and symbols.

So what is the simple logical way to understand that part of the reasoning?

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I think I've identified what confuses me the most. It seems that DeMorgan's law's talks about the equality of two sets in terms of what they have, but the fact that in the question premise we are dealing with the empty set itself (i.e. a set that has nothing) is what is confusing me and then making a jump to a set that has everything coupled with DeMorgan's, complements seems to confusing to deal with all at once. I think because the set has nothing and the proofs I have for those are in terms of things that I do have, is what is confusing me. Perhaps this can bias answers to address what seems to be the crux of my confusion.

Answer 1

If $x \in K_{\alpha}$ for each and every $K_{\alpha}$ then $x \in \cap_{\alpha} K_{\alpha}$.

So if $x \not \in \cap_{\alpha} K_{\alpha}$ then there must exist some $K_f$ where $x \not \in K_f$. So $x \in K_f^c$. So $x \in K_f^c \subset \cup_{\alpha} K_{\alpha}^c$

If $\cap_\alpha K_\alpha = \emptyset$ then for any $x$ in universal space $U$ then $x \not \in \cap_{\alpha} K_{\alpha}$ and hence $x \in K_f^c \subset \cup_{\alpha} K_{\alpha}^c$, for some $K_f$. The is true for all $x \in U$ so $U \subset \cup_{\alpha} K_{\alpha}^c\subset U$. So $\cup_{\alpha} K_{\alpha}^c=U$.

Answer 2

What you call a "symbol crunching game", is a clear and easy step to me. Take any $x\in X$; then $x$ is not in $\cap_\alpha K_\alpha$. That means that there exists (at least) an $\alpha$ with $x\in K_\alpha^c$. In particular, $x\in\bigcup_\alpha K_\alpha^c$.

As we can do the above for any $x$, we conclude that $X\subset\bigcup_\alpha K_\alpha^c$.

Answer 3

Perhaps you need to do quantifiers. Convince yourself that $$ \neg(\exists x\in T) [ F(x)] $$ is logically equivalent to $$ (\forall x \in T) [\neg F(x)] $$

Answer 4

I think I have a nice way to answer this thanks to fleabood and the other provided answers.

"English" version of it:

We note the following things:

1. the intersection of $K_{\alpha}$ is empty
2. Thus, every point in the ambient space $X$ is not in intersection.
3. A point not in the intersection means that it must be in some complement. Thus it must belong to the union of complements of $ K_{\alpha} $
4. Now together with all these, since every point is not in the intersection of $ K_{\alpha}$ and every point not in the intersection of $k_{\alpha}$ is in the complement of some $K^c_{alpha}$, it must mean every point in $X$ is in the union of complements of $K^c_{\alpha}$. Since we can only have points from $X$ and they are all in the union of $K^c_{\alpha}$, it must mean they are the same set.

Now the "symbol version" of it:

1. $\forall x \in X, x \not \in \cap K^c_{\alpha}$ because $ K^c_{\alpha} = \emptyset$
2. for all points not in $\cap K_{\alpha}$ must mean they are in some complement i.e. $x \not \in \cap K_{\alpha} \implies \exists \alpha, x \in K^c_{\alpha} \iff x \in \cup K^c_{\alpha}$

the two together we have $\forall x \in X, x \not \in \cap K_{\alpha}$ AND $\forall x \not \in \cap K_{\alpha} \implies x \in \cup K^c_{\alpha}$ basically remove the middle part that "chains the statements" gives a conclusion about all point in the space. i.e. that they are in the complements of $K^c_{\alpha}$:

$$ \forall x \in X, x \not \in \cap K_{\alpha} \implies x \in \cup K^c_{\alpha} $$

$$ \forall x \in X, x \in \cup K^c_{\alpha} $$

so basically since every point in $X$ is not in the intersection then it must be in a complement and this holds for all $x \in X$ so every point is in $\cup_{\alpha} K^c_{\alpha}$. Since there are no more point to consider it must mean (just to make the other direction very clear too):

$$ \cup_{\alpha} K^c_{\alpha} = X $$

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Summarized version:

A point in the intersection of $K_{\alpha}$ is not in any of the complements. Then we notice that the intersection is empty so no point in $X$ is in the intersection. A point not in the intersection must be in some complement so it must be that all points are in some complement (since all points in X are not in the intersection which means they are in some complement).

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Flea's useful comment:

If the set has nothing in it, then any point that does exist at all.... is not in it. And for a point to not be in an intersection, means the point must not be in at least one of the complements. So each point must be in at least one of the complements. Now different points might belong to different complements but each point belongs to some complement.