Exercise 2.4.8

Question

Regard $\mathbb{R}$ as a subset of $\mathbb{R}^{2}$ in the usual way. Let $\mathscr{T}$ be the family of all subsets $U$ of $\mathbb{R}^{2}$ such that $U \backslash \mathbb{R}$ is an open subset of $\mathbb{R}^{2}$ (in the usual topology). Establish the following assertions:
\begin{enumerate}[label=(\alph*)]
\item $\mathscr{T}$ is a topology for $\mathbb{R}^{2}$.
\item $\left(\mathbb{R}^{2}, \mathscr{T}\right)$ is separable.
\item The subspace $\mathbb{R}$ of $\left(\mathbb{R}^{2}, \mathscr{T}\right)$ is neither separable nor second-countable.
\item $\left(\mathbb{R}^{2}, \mathscr{T}\right)$ is not second-countable.
\end{enumerate}

Elu Thingol · Accepted Answer

The definition of $\mathcal{T}$ as it stands is incorrect; we take $\mathcal{T}$ to consist of all sets of the form $U=V \backslash A$, where $V$ is an open subset of the $\mathbf{R}^{2}$ (in the usual topology), and $A$ is an arbitrary subset of $\mathbf{R}$.

\begin{proof}
\begin{enumerate}[label=(\alph*)]
\item 	extit{$\mathscr{T}$ is a topology for $\mathbb{R}^{2}$.}
ewline
The fact that $\mathcal{T}$ is a topology follows from the fact that we can interchange set differences and unions/intersections.

\item 	extit{$\left(\mathbb{R}^{2}, \mathscr{T}ight)$ is separable.}
ewline
Let $z=(x, 0) \in \mathbf{R}$. For each $r>0$, the set $(B(z, r) \backslash \mathbf{R}) \cup\{z\}$ is a $\mathcal{T}$-open set containing $z$, and every $\mathcal{T}$-open set containing $z$ contains a set of this form. From this, it follows easily that the points with rational coordinates are $\mathcal{T}$ dense so that the space is separable.

\item 	extit{The subspace $\mathbb{R}$ of $\left(\mathbb{R}^{2}, \mathscr{T}ight)$ is neither separable nor second-countable.}
ewline
The relative $\mathcal{T}$-topology for $\mathbf{R}$ is the discrete topology, which is neither separable nor second-countable.

\item 	extit{$\left(\mathbb{R}^{2}, \mathscr{T}ight)$ is not second-countable.}
ewline
In view of \href{./exercise_2-4-7}{Exercise 7} and the previous part, the topology $\mathcal{T}$ is not second-countable.
\end{enumerate}
\end{proof}

Exercise 2.4.8

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Exercise 2.4.8

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