I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Prove that any intersection or finite union of closed sets is closed.
My Proof:
(Intersection)
Let A = ⋂F where F is a family of n closed sets F = {X1,…,Xn}. Let (xm) be a convergent sequence completely contained in A. This means that (xm) is also completely contained in Xk for all sets in F, and since each set is closed, lim(xm) is an element of every set in F. Since lim(xm) is an element of every set of F, it is also an element of A. Thus any convergent sequence completely contained in an intersection of closed sets also has its limit contained in that intersection. So any intersection of closed sets is closed.
(Finite Union)
Now let B = UF, and let (xm) be a convergent sequence completely contained in B. Since (xm) is convergent, it is Cauchy and thus any neighborhood around lim (xm) will contain all but finitely many elements of (xm). Since (xm) is contained in UF, there must then be some Xj∈F which contains all but finitely many elements of (xm)*. Define (ym) as a subsequence of (xm) which only contains values contained in Xj. We can see that (ym) is a convergent sequence contained in Xj, and since Xj is closed, lim(ym)∈Xj. The limit of (xm) is the same as the limit of any of its convergent subsequences, so lim(xm)∈Xj and thus lim(xm)∈B. Thus, any convergent sequence in a union of closed sets has its limit contained in that union of closed sets, so a finite union of closed sets is closed.
*The part in bold seems a little hand-wavy to me. Is that justification enough, or is there some step I am missing?
Best Answer
The first proof is great - no problems there! For the second proof, you have the right idea, but the bold statement does require proof. Luckily it's true because of the Pigeonhole Principle.
Colloquially, "if you stuff infinitely many pigeons into finitely many holes, at least one hole will have infinitely many pigeons." More formally, suppose $B = \cup \, X_i$, where each $X_i$ is closed. By definition, the sequence $\{x_m\}$ has infinitely many terms, so at least one $X_i$ -- without loss of generality, call it $X_1$ -- must contain infinitely many terms of the sequence. In other words, $X_1$ contains a sub-sequence.
Then as you did, you can use the fact that $X_1$ is closed to show that $\lim x_m$ is contained in $X_1$, and thus contained in $B$. Hope that helps!