Exercise 3.4.5 of Avner Friedman's Foundations of Modern Analysis asks the following question
A set $Y$ in metric $X$ is first category in $X$ if it is contained in a countable union of nowhere dense sets of $X$. If $Y$ is not first category, it is second category. The real line with Euclidean metric is a space of the 2nd category. Prove, however, that as a subset of the Euclidean plain, the real line is a set of first category.
And here is my summary of why this true, and where I get stuck.
$R^1$ is second category clearly since it is a complete metric space and all complete metric spaces are 2nd category. Now consider $R^1\subseteq R^2=\{(a,b): a,n\in R\}$ where $R^1=\{(a,0): a\in R\}$. We want to show that $R^1$ is contained in a countable union of nowhere dense sets of $R^2$. That is, we want to show $R^1\subseteq S=\bigcup_{j=1}^\infty T_j$ where $T_j$'s are nowhere dense, i.e. $\overline{T_j}$ has no interior points.
Essentially what we need to see is that $R^1$ is not complete in $R^2$, meaning that we can find a Cauchy sequence in $R^1$ that has its limit in $R^2- R^1$.
So then I started thinking about this and started losing all hope for any understand that I had about what it means for a set to be dense verses nowhere dense. I could really use some help to see how to finish this proof, or help on really understanding these two concepts and how they are different. Thanks a bunch!
Best Answer
$\mathbb{R}$ is nowhere dense in $\mathbb{R}^2$, as it is closed in $\mathbb{R}^2$ and has an empty interior. In particular, it is of the first category (if you want a countably infinite union of nowhere dense sets, take $\mathbb{R}$ together with the empty set countably many times).