It's more of a conceptual question. The discrete topology, I understand, is essentially the power set, so every possible subset put together for a set $X$.
But when it comes to $\mathbb{R}$…the elements are infinite. There are infinitely many reals within any interval, like $0$ to $1$. The Euclidean topology is induced by the Euclidean metric, and so basically we say all elements within a certain "distance" is open. So take any $x<y \in \mathbb{R}$ and any $c$ that $x<c<y$ is in an open set and namely, in the topology. Since $x,y$ are arbitrary…does this give the power set of $\mathbb{R}$?
How about in the general $\mathbb{R}^n$? I cannot find a sensible way to describe the power set of the reals apart from this, using the Euclidean topology. I mean, it seems to contain all possible subsets of the reals.
An elaborate explanation would be very much appreciated.
Best Answer
A point $x \in X$ is said to be near a set $Y \subseteq X$, if every neighborhood of $x$ intersects $Y$.
In the standard (Euclidean metric) topology for $\Bbb R^n$, the near points of an open ball of radius $r< 1$ centered at $a$ are the elements of the open ball, and its boundary (so, a closed ball centered at $a$ of radius $r$). This is loose enough so we get "plenty" of near points, but exclude anything "not too close".
If we take the discrete topology on $\Bbb R^n$ (induced by the discrete metric), an open ball of radius $r < 1$ centered at $a$ contains "just $a$". In other words, in the discrete topology , every other point but $a$ is "far, far away" from $a$.
That is to say, the usual topology and the discrete topology behave qualitatively differently.
As for your question about "size", it turns out that the size of the Euclidean topology on $\Bbb R$ is the same size as $\Bbb R$, whereas the power set is "bigger" (the proof of my first statement rests upon the fact that we can write any open interval as a countable union of open intervals with rational endpoints, using Cauchy sequences, the second statement is due to Cantor, who showed there is no bijection between $X$ and $2^{X}$ for any set $X$).