I have been reading this text http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF, and before I address my main question, I want to note that the author, in the same section, mentioned that for the complement of any set $S$ we are always considering the elements that are not in $S$ but in the universal, that is the reals:
Now, the proof that the intersection of a closed set is closed is given as:
The only problem I have with understanding proof (b) is that the complement of $T$, in this context, is the set of all numbers that are not members of $T$ but members of the reals, but, we are told that the complement of $T$ is the union of the complements of all of the sets $F$ which would infer that there is no element of at least one $F$ that is not an element of $T$. Yet, this is a contradiction to $T$ being an intersection of the sets $F$, as it states that $T$ is equal to all of the $F$s, or in other words, $T$ contains every element of every $F$. So, my question is:
In this context, is the complement of $T$ still defined in terms of the reals or is it defined relatively $T$? For, if it is defined relatively, then I can progress with the proof since I know the proof for the relative complement of an intersection.
(Note: I had a look at this question Proof of complement of intersection defined using an arbitrary set. and it seems that the complement is defined relatively. Also, sorry if this question doesn't make sense but this particular part of the proof just doesn't make sense to me at all.)
Best Answer
I seem to understand that the problem is the equality $\left( \bigcap A_n\right)^C=\bigcup A_n^C$. So let's take an element $x\in \left( \bigcap A\right)^C$ then there is an $A_n$ such that $x\notin A_n$ which means $x\in A_n^C$ and then $x\in \bigcup A_n^C$. Now let's take $x\in \bigcup A_n^C$ then there is some $A_n$ for which $x\notin A_n$ and then $x\in \left( \bigcap A\right)^C$