Exercise 16.8

Question

If $L$ is a straight line in the plane, describe the topology $L$ inherits as a subspace of $\mathbb{R}_\ell 	imes \mathbb{R}$ and as a subspace of $\mathbb{R}_\ell 	imes \mathbb{R}_\ell$. In each case it is a familiar topology.

Amit Awasthi · Accepted Answer

\begin{proof}[Solution]
  Note that the basis for $\mathbb{R}_\ell 	imes \mathbb{R}$ consists of
  elements of the form $[a,b) 	imes (c,d)$. If $L = \{(x,y) \mid x = x_0\}$,
  then $L \cap [a,b) 	imes (c,d) = \emptyset$ or $\{x_0\} 	imes (c,d)$, and
  so defining the map $\varphi\colon L \cap (\mathbb{R}_\ell 	imes \mathbb{R}) 	o \mathbb{R}, \{x_0\} 	imes (c,d) \mapsto (c,d)$, it is bijective, open, and continuous, and so the topology $L$ inherits is homeomorphic to $\mathbb{R}$ with the standard topology. If $L$ has finite slope, we first note that $L \cap (\mathbb{R}_\ell 	imes \mathbb{R}) = \{(x,mx+b) \in \mathbb{R}^2 \mid x \in \mathbb{R}\}$, and that the basis for our topology are the sets of the form $\emptyset, [ (a,ma+b),(c,mc+b) ), ( (a,ma+b),(c,mc+b) )$ for $a,c \in \mathbb{R}$ and $a < c$, by Lemma $16.1$. We then define
  \begin{equation*}
    \varphi\colon L \cap (\mathbb{R}_\ell 	imes \mathbb{R}) 	o
    \mathbb{R}_\ell, \quad (a,ma+b) \mapsto a.
  \end{equation*}
  This implies
  \begin{align*}
    ( (a,ma+b),(c,mc+b) ) &\mapsto (a,c),\
    [ (a,ma+b),(c,mc+b) ) &\mapsto [a,c).
  \end{align*}
  We claim this defines a homeomorphism with $\mathbb{R}_\ell$. Clearly, it is continuous, for the basis elements of $\mathbb{R}_\ell$ have preimages that are basis elements in the topology on $L$. Likewise, it is open since the basis elements of $L$ map to sets that are open in $\mathbb{R}_\ell$ by Lemma $13.4$. Finally this is a bijection since there exists an inverse just by reversing the arrows above.
  \par For $\mathbb{R}_\ell 	imes \mathbb{R}_\ell$, by following the same steps
  as above if $L = \{(x,y) \mid x = x_0\}$, then $L \cap (\mathbb{R}_\ell 	imes
  \mathbb{R}_\ell)$ is homeomorphic to $\mathbb{R}_\ell$. For $L$ with $|m| <
  \infty$, we must split it up into two cases. When $m \ge 0$, we have a similar situation as above, except we only have to consider basis elements of the form $[a,b)$; thus, $L \cap (\mathbb{R}_\ell 	imes \mathbb{R}_\ell)$ is homeomorphic to $\mathbb{R}_\ell$. When $m < 0$, since for every point $(x,y) \in L$, we can find a basis element $[x,a)	imes[y,b) \in (\mathbb{R}_\ell 	imes \mathbb{R}_\ell)$ such that $L \cap [x,a)	imes[y,b) = \{(x,y)\}$, and these form the open sets of our new topology by Lemma $16.1$. We see then that the topology on $L$ is homeomorphic to the discrete topology on $\mathbb{R}$.
\end{proof}

Dan Whitman · Answer

\def\Rl{${\mathbb{R}}_l$\,}
\def\Ru{${\mathbb{R}}_u$\,}
\def\Rd{${\mathbb{R}}_d$\,}

First, let \Ru denote the reals with the upper limit topology, with a basis containing all intervals $(a,b]$ for $a < b$.
Also let \Rd denote the reals with the discrete topology, which can clearly be generated by a basis containing intervals $[a,b]$ for $a \leq b$.
This is easy to see as $[a,a] = \left\{aight\}$ is a basis element so that any subset of ${\mathbb{R}}$ can be considered a union of such basis elements.
It is then easy to show that \Rl and \Ru are both strictly finer than the standard topology on ${\mathbb{R}}$ (this was shown in Lemma~13.4 for \Rl), but that \Rl and \Ru are incomparable.
Clearly, \Rd is strictly finer than both of these since it is the finest possible topology on ${\mathbb{R}}$.

Now, regarding the main problem, we do not yet have the tools show formally show how topologies on a line $L$ compare to topologies on ${\mathbb{R}}$, so we will have to discuss this informally.
We can see that, in some sense, a line $L$ in the plane is like a copy of the real line so that we can discuss topologies on $L$ as being in some sense the same as topologies on ${\mathbb{R}}$.

The product topology ${\mathbb{R}}_l 	imes {\mathbb{R}}$ is the topology generated by the basis containing sets of the form $[a,b) 	imes (c,d)$ where $a < b$ and $c < d$ by Theorem~15.1.
Then, for a line $L$ in the plane, it can intersect such a basis element in a variety of ways, which are illustrated below:

\def\fsize{3cm}
\def	ikzgray{black!30!white}

ewcommand\linebe[2]{
\def\lbb{0.2}
\begin{tikzpicture}[scale=3]
% Draw rect to ensure the space is used
\fill[fill=white,opacity=0] (0,0) rectangle (0,0);

% Basis element shape
\fill[fill=	ikzgray] (\lbb,\lbb) rectangle (1-\lbb,1-\lbb);
\draw[thick] (\lbb,\lbb) -- (\lbb,1-\lbb);
\draw[thick,dashed] (\lbb,1-\lbb) -- (1-\lbb,1-\lbb) -- (1-\lbb,\lbb) -- (\lbb,\lbb);

% Line
\draw[thick] (#1) -- (#2);
\end{tikzpicture}
}
\begin{figure}[H]
	ext{(a)} \linebe{0,0.5}{1,0.5} \hspace{0.4cm} 	ext{(b)} \linebe{0.5,0}{0.5,1} \
	ext{(c)} \linebe{0,0.2}{0.8,1} \hspace{1cm} 	ext{(d)} \linebe{0.20,0}{1,0.8} \
	ext{(e)} \linebe{0,0.8}{0.8,0} \hspace{1cm} 	ext{(f)} \linebe{0.2,1}{1,0.2}
\end{figure}

Clearly, the intersection of $L$ and these basis elements results in some kind of interval on $L$.
Such intervals then form the basis for the subspace topology on $L$ by Lemma~16.1 since they are the intersection of $L$ and a basis element in the superspace.
Another point is that the orientation of the line $L$ with regard to the way in which it is a copy of ${\mathbb{R}}$ is important.
For example, in Figure~(a) above, if $L$ is oriented in a natural way with the negative reals on the left and the positive reals on the right, then the resulting intervals are of the form $[a,b)$, which would result in a topology like \Rl.
The opposite orientation results in intervals of the form $(a,b]$ as basis elements, generating a topology like \Ru.

Now, for a line $L$ such as that illustrated in Figure~(a), every possible basis element of ${\mathbb{R}}_l 	imes {\mathbb{R}}$ that intersects $L$ results in the half-open intervals as described above depending on the orientation of $L$.
This is not the case for all lines, however, and is dependent on their slope in the plane.
For example, lines with positive slope can intersect basis elements as in Figure~(c), which results in half-open intervals $[a,b)$ (or $(a,b]$ depending on orientation), or they can intersect them as in Figure~(d), which results in open intervals $(a,b)$.
However, since the topologies \Rl and \Ru are strictly finer than the standard topology, the subspace topology formed on $L$ would be like these (which depends on orientation) rather than like the standard topology.
Lastly, we note that, for any appropriate interval on the line $L$, we can clearly always find a basis element $B$ in ${\mathbb{R}}_l 	imes {\mathbb{R}}$ such that the intersection of $B$ with $L$ is the interval.
For this reason, these intervals form the basis elements of the topology on $L$.

With all these considerations in mind, we list the topologies on ${\mathbb{R}}$ that the subspace topologies on $L$ are like based on line directions and orientations for product topologies ${\mathbb{R}}_l 	imes {\mathbb{R}}$ and ${\mathbb{R}}_l 	imes {\mathbb{R}}_l$:
\begin{center}
\begin{tabular}{c|cc}
$L$ & ${\mathbb{R}}_l 	imes {\mathbb{R}}$ & ${\mathbb{R}}_l 	imes {\mathbb{R}}_l$ \
\hline
$ightarrow$ & \Rl & \Rl \
$
earrow$ & \Rl & \Rl \
$\uparrow$ & ${\mathbb{R}}$ & \Rl \
$
warrow$ & \Ru & \Rd \
$\leftarrow$ & \Ru & \Ru \
$\swarrow$ & \Ru & \Ru \
$\downarrow$ & ${\mathbb{R}}$ & \Ru \
$\searrow$ & \Rl & \Rd
\end{tabular}
\end{center}
We note that ${\mathbb{R}}$ simply denotes the standard topology.

$L$	$ℝ_{l} \times ℝ$	$ℝ_{l} \times ℝ_{l}$

$\to$	$ℝ_{l}$	$ℝ_{l}$
$↗$	$ℝ_{l}$	$ℝ_{l}$
$↑$	$ℝ$	$ℝ_{l}$
$↖$	$ℝ_{u}$	$ℝ_{d}$
$\leftarrow$	$ℝ_{u}$	$ℝ_{u}$
$↙$	$ℝ_{u}$	$ℝ_{u}$
$↓$	$ℝ$	$ℝ_{u}$
$↘$	$ℝ_{l}$	$ℝ_{d}$

Exercise 16.8

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Exercise 16.8

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