To show that $T$ is a topology, indeed, you need to show the following three things:-
$\emptyset, X \in T$.
For an arbitrary collection $\mathscr{F} = \left\lbrace G_{\lambda} | \lambda \in \Lambda \right\rbrace$ from $T$, the union $\bigcup\limits_{\lambda \in \Lambda} G_{\lambda}$ is in $T$.
For two sets $G_{1}, G_{2} \in T$, the intersection $G_1 \cap G_2 \in T$.
To show the first point, let us first observe a point. Suppose we have a family $\mathscr{F}$ of sets, which is indexed by an index set $\Lambda$. Note that this index set may be finite, countable or uncountable, i.e., we do not know the nature of $\Lambda$, but only know that it indexes the sets in $\mathscr{F}$.
Now, consider the family of open balls $\mathscr{F} = \left\lbrace B \left( {x_{\lambda}, r_{\lambda}} \right) | \lambda \in \emptyset \right\rbrace$. Therefore, this family is indexed by the empty set. As out intuition tells us, the family must be empty (since there annot be any $\lambda \in \emptyset$ and so shall be its union. We shall prove this assertion as follows:
Let, if possible, $\bigcup\limits_{\lambda \in \emptyset} B \left( x_{\lambda}, r_{\lambda} \right) \neq \emptyset$. Then, there is an element, say $y$, in this union. In particular, there is some $\lambda_0 \in \emptyset$ such that $y \in B \left( x_{\lambda_0}, r_{\lambda_0} \right)$. However, this is not possible. Hence, the union must be empty.
Thus, we have proved that empty set can be written as a union of open balls (empty union) so that $\emptyset \in T$, by the definition of $T$.
Now, consider for each $x \in X$, the open ball $B \left( x, 1 \right)$. Clearly, $x \in B \left( x, 1 \right)$ so that if we consider the union of all such open balls as $x$ varies over $X$, we must get the whole space $X$. Now, we will prove this assertion.
Already, we know that $\bigcup\limits_{x \in X} B \left( x, 1 \right) \subseteq X$, since $X$ is the universal set. Now, let $y \in X$. We know that $y \in B \left( y, 1 \right) \subseteq \bigcup\limits_{x \in X} B \left( x, 1 \right)$ so that $X = \bigcup\limits_{x \in X} B \left( x, 1 \right) \in T$.
For the next two steps, we will use the property of open sets in a metric space. Since $\left( X, d \right)$ is a metric space, we have the following properties at hand.
- Any open set in $X$ is the union of open balls.
- Arbitrary union of open sets is open.
- Intersection of two open sets is open.
Now, to prove point (2) for topology $T$, we shall employ points (1) and (2) of the properties of open sets in $X$. Taken any arbitrary collection of union of open balls, it is the same as an arbitrary collection of open sets. Hence, its union is open and again, by property (1), it must be a union of open balls. Therefore, arbitrary unions of sets from $T$ are again in $T$.
Similarly, intersection of two unions of open balls is an open set in $X$ (metric space) so that it is a union of open balls, therefore belonging in $T$.
Hence, we have proved that $T$ is indeed a topology on $X$.
As for answering about your grasp on $T$, I suggest you to note the following things. When we first developed the concept of metric space, we did it to "measure" the "distance" between two points in any given sets. Immediately, we observed that one implication of defining distance was that we were getting something, which we called "open sets" and using these we further defined properties such as convergence of sequences, continuity of functions, compactness, connectedness and others.
However, we soon realised that everytime defining some "metric" is tiresome and indeed imposes some restrictions. Therefore, we wanted to abstract the important point of metric spaces so that we could define all the things we did in an analogous way. Since in the definitions (of those mentioned above), we have used open sets as a major source of information, mathematicians then tried to abstract open sets themselves.
Therefore, now instead of defining a metric and then creating open sets through the metric, let us directly define a collection of sets from our universal set, which we shall call "open". However, simply taking any sets to be open will not work. This is because, when we abstract certain ideas, the basic properties should remain. Therefore, we must ask: "What are the basic properties of open sets?"
Indeed, we get the three points that define a topology as the answer to this question.
Hence, whenever you want to prove/disprove that a given collection of sets defines a topology on a set $X$, all you need to do is to check these three properties. Most of the times, the way in which $T$ is defined (which is usually the set-builder form, and not mere listing of elements) gives hints and infact, a way to proceed with the proof. However, if the sets in $T$ are merely listed, you will need to check literally every combination possible.
Best Answer
After rearrangement we need to prove the following:
$\frac{|A \cap B |}{|A \cup B|} + \frac{|B \cap C|}{|B \cup C|} \leq 1 + \frac{|A \cap C| }{|A \cup C |}$
For notational purposes, define $a = |A \setminus (C\cup B)|, c = |C \setminus (A\cup B)|, d = |(A \cap C) \setminus B|.$ Further define $b_1 = |B \cap (A \setminus C)|, b_2 = |B \cap (C\setminus A)|, b_3 = |B \cap A \cap C|, b_4 = |B \setminus (A\cup C)|$. Then the above inequality can be expressed as
$\frac{b_1 + b_3}{a+d+b_1 + b_2 +b_3+ b_4} + \frac{b_2 + b_3}{c+d+b_1 +b_2+b_3 + b_4} \leq 1 + \frac{d+b_3}{a+c+d+b_1 +b_2 +b_3} $
So we need to prove validity of this algebraic expression. With the last notational definition of $b = b_1 + b_2 +b_3$ we get by inspection
$LHS = \frac{b-b_2}{a+d+b+b_4} + \frac{b-b_1}{c+d+b + b_4} \leq \frac{b-b_2}{a+d+b} + \frac{b-b_1}{c+d+b} = b[\frac{1}{a+d+b} + \frac{1}{c+d+b}] - \frac{b_1}{c+d+b} - \frac{b_2}{a+d+b}$
$RHS = 1 + \frac{d+b-(b_1 + b_2)}{a+c+d+b} =( 1+ \frac{d+b}{a+c+d+b}) - \frac{b_1}{a+c+d+b} - \frac{b_2}{a+c+d+b}$
Immediately we see that the negative parts obey the inequality. Hence check if the following holds:
$b(\frac{1}{a+d+b} + \frac{1}{c+d+b}) \leq 1 + \frac{d+b}{a+c+d+b} $
which is equivalent to
$b(2+ \frac{c}{a+d+b} + \frac{a}{c+d+b}) \leq 2(b+d) + c+a $
and we see that the last inequality is true since $\frac{b}{a+d+b} \leq 1, \frac{b}{c+d+b} \leq 1$. Therefore $LHS \leq RHS$ which is what we wanted to prove.