Alternative proof to Rudin's "any intersection of closed sets is closed" I was trying to proove Theorem 2.24b) in Rudin. I saw its proof which makes sense to me and has been re-phrased in this website and the web multiple times it seems to me. However, I had thought of a different proof and wanted to see if it was correct (which I suspect there is one fishy step in it) and will point it out:
Proof: we want to show any collection of closed sets when intersected is also closed. So we want to show $ \cap_{\alpha} F_{\alpha}$ is closed i.e. contains all its own limit points. With some abuse of notation we can say we want to show: 
$$ \cap_{\alpha} F_{\alpha} = \cap_{\alpha} F_{\alpha} \cup (\cap_{\alpha} F_{\alpha})'$$
 where ' denotes limit points (better notation would be $\bar A =A \cup A'$ but writing bars gets too messy).
Using this same abuse of notation we notice $F_{\alpha} = F_{\alpha } \cup F_{\alpha}'$ which must means $\cap_{\alpha} F_{\alpha} = \cap_{\alpha} ( F_{\alpha} \cup F_{\alpha}')$. Intuitively, we are intersecting all the (normal) sets $F_{\alpha}$ and its limit points also $F_{\alpha}'$. So we either gets intersections of $F_{\alpha}$ or intersections of its limit points. This must mean (which is the step I think is wrong):
$$\cap_{\alpha} F_{\alpha} = (\cap_{\alpha} F_{\alpha} ) \cup ( \cap_{\alpha} F_{\alpha}') )$$
now we just need to show $\cap_{\alpha} F_{\alpha}'$ is the limit points of the intersection we want and we are done. i.e. show $(\cap_{\alpha} F_{\alpha})' = \cap_{\alpha} F_{\alpha}'$. We notice that if $x \in \cap_{\alpha} F'_{\alpha}$ then $x$ is a limit point of all the $F_{\alpha}$'s simultaneously. This is exactly what being a limit point of the intersection means, so we showed that $x$ is a limit point of $ \cap_{\alpha} F_{\alpha}$. Thus $(\cap_{\alpha} F_{\alpha})' = \cap_{\alpha} F_{\alpha}'$. 
Is this proof correct?

Reason I think that step is wrong:
The reason I think that step is wrong is due to commutativity of unions. Check the simple example:
$$ (A \cup B) \cap (C \cup D) = (A \cap C) \cup (B \cap D) $$
this seems right if you draw it but if you switch the C and D around and do it again, the picture doesn't seem quite right anymore I think...also using an analogy to multiplication being intersection and addition being union, the statement is false cuz there are some cross terms missing...but what I wrote intuitively:

Intuitively, we are intersecting all the (normal) sets $F_{\alpha}$ and its limit points also $F_{\alpha}'$. So we either gets intersections of $F_{\alpha}$ or intersections of its limit points.

seems extremely plausible. So I am not sure where my mistake is...or if there is any. Sometimes the natural language translating to formal is weird.

Side note:
I guess in a more philosophical perspective I don't understand why all the other proofs use De Morgan's or why thats the natural proof to think of. For me Rudin's proof seemed link symbol crunching (mechanicaly) so I am not sure if it would have occurred to me but with its high frequency in the web I assume this is what people usually see (seems like a parrot proof where people just repeat it cuz its what everyone knows, just my hunch). Anyway, I don't get it, why is that proof the more common proof? Is it because its the only known proof to be correct? Or why? What was the intuition, thought process that went into creating Rudin's proof that uses De Morgan's and Complements of open sets are closed sets? Is it that he just enumerated all the facts he new up till now and combined them until it worked? Its just the way it feels to me because those facts where presented in before thm 2.24 and then by coincidence used to proof it. Just trying tog et some insight into the creation of mathematics.
So the two things I don't think I would have thought of:


*

*Use DeMorgan's (just why?)

*since I wouldn't have thought of 1) I obviously didn't see it coming that complement of open sets being closed would eventually be useful.



For reference the related questions/linls:


*

*Arbitrary intersection of closed sets is closed

*Proving intersection of closed sets is closed (with a specific definition)

*https://proofwiki.org/wiki/Intersection_of_Closed_Sets_is_Closed
 A: My metric-space-topology definition of a closed set is a set containing all its adherent points (this is equivalent to your definition of "a set containing all its limit points").
Proof: Let $x_0$ be an adherent point of $\cap_\alpha F_\alpha$. By definition, for every $r > 0$, the open ball $B(x_0, r)$ has non-empty intersection with $\cap_\alpha F_\alpha$, i.e., $\exists y \in \cap_\alpha F_\alpha, d(x_0, y) < r$; this shows that for all $\alpha$, $B(x_0, r)$ also has non-empty intersection with $F_\alpha$ (because $y \in F_\alpha$), and therefore $x_0$ is an adherent point of $F_\alpha$, then $x_0 \in F_\alpha$ (because $F_\alpha$ is closed). This allows us to conclude that $x_0 \in \cap_\alpha F_\alpha$, and thus $\cap_\alpha F_\alpha$ is closed.
A: the step you think is wrong is wrong and for the reason you think it is wrong.  
Consider four arbitrary sets $A,B,C,D$.   We can break the universe into $2^4 = 16$ disjoint subsets by considering the sets that contain all the points that are (or are not) in those four sets.
Then $(A\cup B) \cap (C \cup D)$ will consist precisely of the follow sets:


*

*$AB'CD'$: This is the set of all elements that are in $A$; not in $B$; in $C$; not in $D$
- $AB'C'D$:This is the set of all elements that are in $A$; not in $B$; not in $C$;  in $D$

*$AB'CD$:This is the set of all elements that are in $A$; not in $B$;  in $C$;  in $D'$

*$ABCD'$: This is the set of all elements that are in $A$; in $B$; in $C$; not in $D$

*$ABC'D$:This is the set of all elements that are in $A$; in $B$; not in $C$;  in $D$

*$ABCD$:This is the set of all elements that are in $A$;  in $B$;  in $C$;  in $D'$
- $A'BCD'$: This is the set of all elements that are not in $A$; in $B$; in $C$; not in $D$

*$A'BC'D$:This is the set of all elements that are not in $A$;  in $B$; not in $C$;  in $D$

*$A'BCD$:This is the set of all elements that are not in $A$;  in $B$;  in $C$;  in $D'$


Whereas $(A\cap C)\cup(B\cap D)$ consist of:


*

*$AB'CD'$: This s the set of all elements that are in $A$; not in $B$;  in $C$; not in $D$. 

*$ABCD'$: This s the set of all elements that are in $A$;  in $B$;  in $C$; not in $D$. 

*$AB'CD$: This s the set of all elements that are in $A$; not in $B$;  in $C$;  in $D$.

*$ABCD$: This s the set of all elements that are in $A$;  in $B$;  in $C$;  in $D$.  

*$A'BC'D$: This s the set of all elements that are not in $A$;  in $B$;  not in $C$;  in $D$. 

*$A'BCD$: This s the set of all elements that are not in $A$;  in $B$;  in $C$;  in $D$. 

*$ABC'D$: This s the set of all elements that are in $A$;  in $B$; not in $C$;  in $D$.


These are different groups of sets.   In particular the bold ones are in one but not the other group.
So we can see that $(A\cap C)\cup(B\cap B) \subsetneq (A\cup B)\cap(C \cup D)$ and we can prove it by element chasing:
If $x\in(A\cap C)\cup(B\cap D)$ then either i)$x\in A\cap C$ in which case $x\in A \subset (A\cup B)$ and $x \in C\subset (C\cup D)$; or ii) $x \in B\cap D)$ in which case $x \in B\subset (A\cup B)$ and $x \in D\subset(C\cup D)$.  Either way $x$ is in both $(A\cup B)$ and $C\cup D$ so $x\in (A\cup B) \cap (C\cup D)$.
But if $a \in A\cap C$ but in neither $D$ nor $B$......
Of ccourse drawing a Venn diagram might be the easiest.
A: Indeed the part you suspected to be wrong is wrong.
If $A_n,B_n$ are sequences of closed sets, it is not true that $\bigcap_n (A_n\cup B_n) = \bigcap_n A_n \cup \bigcap_n B_n$
As a counterexample, consider $A_n = [0,\frac{1}{2n}]$ and $B_n=[\frac{1}{2n},1]$.
So $A_n\cup B_n = [0,1]$ hence the left hand side is $[0,1]$.
On the other hand, $\bigcap_n A_n = \{0\}$ and $\bigcap_n B_n = B_1 = [\frac{1}{2},1]$
(If this should be a comment let me know)
