Why was the concept of first/second categories in metric spaces introduced? Let $(X, d) $ be a metric space. $X$ is of first category if it can be expressed as a countable union of nowhere dense subsets. Otherwise, $X$ is called a metric space of second category.
A first category set is intuitively "small " and a second category set is "large". A second category set with no isolated point is uncountable.
Cardinality is purely a set-theoretic characterisation and first/second category is related to the topology on the underlying set.
However I am interested to know whether there are any deeper relations between the cardinality and whether it is first or second category.
Why was the concept of first/second categories in metric spaces introduced?
 A: The Baire category theorem is an abstraction of the following sort of construction, which seems to arise fairly often. You want to construct an object $X$ (of some given sort) that satisfies an infinite list of requirements, say R1, R2, .... Instead of attacking all these requirements at once, you try to handle them one at a time, and you notice (with considerable happiness) that each requirement is amazingly nice, in the sense that, given any object $Y$, you can slightly modify it to satisfy any single R$n$. So you try starting with some arbitrary $Y$, modifying it to get some $Y'$ that satisfies R1, modifying that to get $Y''$ satisfying R2, and --- OOPS: $Y''$ no longer satisfies R1; the second modification ruined what the first had achieved.
But then you notice an improvement on your original happy observation. The R$n$'s are super-nice in the sense that any $Y$ can be modified to a $Y'$ that satisfies R$n$ and continues to satisfy R$n$ when modified sufficiently slightly.
This means that you can modify $Y$ to some $Y'$ that satisfies R1 and this achievement won't be ruined by future modifications if those modifications are slight enough. Now modify $Y'$ very slightly to get $Y''$ satisfying R2 and still satisfying R1, with room to spare. Here "room to spare" means that further, sufficiently slight modifications won't ruin R1 or R2. Continue in this way, producing $Y'''$ and so forth. After $n$ steps, you have $Y^{(n)}$ satisfying R1 through R$n$ and guaranteed to still do so if perturbed extremely slightly. (At each step, modify the previous $Y^{(n)}$ much less than any of your previous $n-1$ modifications.)
OK, so you can satisfy any finitely many of your requirements with a suitable $Y^{(n)}$, but you want to satisfy all infinitely many requirements with a single object $X$. Well, if you've made the modifications in your construction slight enough, the sequence $Y, Y', Y'',\dots,Y^{(n)},\dots$ will be a Cauchy sequence. If the objects you're working with constitute a complete space, then this sequence will converge to an $X$ that is close enough to each $Y^{(n)}$ to guarantee that $X$ satisfies R$n$. So that limit $X$ is the object you want.
The key ingredients in this argument are (1) completeness of the space; (2) availability of a set $G_n$ of objects that satisfy R$n$ and continue to do so when sufficiently slightly modified, i.e., an open set of objects satisfying R$n$; and (3) the possibility of extremely slightly modifying any object to get it into $G_n$. In other words, $G_n$ is dense in the whole space of your objects.
So, isolating the key ingredients, we have the theorem: In a complete metric space, the intersection of any countably many dense open sets is nonempty.
The complement of a dense open set is a nowhere dense closed set, so the theorem is equivalent to: A complete metric space is not the union of countably many nowhere dense (closed) sets. (I put "closed" in parentheses because omitting it doesn't affect the meaning; a set is nowhere dense iff its closure is nowhere dense.) In other words, a complete metric space is of second category.
