# Is the Euclidean=usual=standard topology on $\mathbb{R}^n$ kind of like the discrete topology?

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.

• Every non-empty open subset of $\mathbb{R}^n$ is uncountable. Apr 29, 2016 at 20:15
• If the Euclidean topology were the discrete topology, then $\{0\}$ would be open, but it isn't because the distance you mention must be positive. Apr 29, 2016 at 20:22

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 $a$.
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 $X$).