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It’s problem #1 of Chapter 2 in Pugh’s book. I think I got the first part by taking $r=1$ and pointing out that $d(p,q) < 1 \implies q \in (0,1)$ for all $p \in (0,1)$ but I have no idea how to show it isn’t a subset of $\mathbb{R}^2$. I tried proof by contradiction and got no where. I’ve been working at this problem for hours and not getting anywhere. How should I do it?

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Your argument for the first part is incorrect: $d(p,q)<1$ and $p\in(0,1)$ do **not** imply that $q\in(0,1)$. For example, take $p=\frac12$ and $q=\frac54$; then $d(p,q)=\frac34<1$, but $\frac54\notin(0,1)$. Try this instead: if $p\in(0,1)$, let $r=\min\{p,1-p\}$, and show that if $d(p,q)<r$, then $q\in(0,1)$. (Note that $r$ is simply the distance from $p$ to the nearer endpoint of $(0,1)$.

For the second part, observe that if $B$ is any open ball whatsoever centred at a point of $(0,1)\times\{0\}$, there are points of $B$ with positive $y$-coordinates; thus, $B$ cannot be a subset of $(0,1)\times\{0\}$. And if $(0,1)\times\{0\}$ contains no open ball, it can’t be open in $\Bbb R^2$. (In particular, if $B$ is the open ball of radius $r$ centred at $\langle x,0\rangle$, where is $\left\langle x,\frac{r}2\right\rangle$?)

If $x\in (0,1)\times \{0\}=L$, can you show that $B(x,\epsilon)$ is not contained in $L$ for any $\epsilon >0$? Note the ball is now a circle, not an open line segment!

To prove it is open in $\Bbb R\times \{0\}\subseteq \Bbb R^2$ note any segment $(a,b)\times \{0\}$ is of the form $B(x,\epsilon)\cap(\Bbb R\times \{0\})$ for a suitable $x$ and $\epsilon >0$.

Recall that $G$ is open in $\Bbb R^2$ iff for each point in $G$ there is an open ball around such point contained in $G$. On the other hand, $G’$ is (relatively) open in $S\subseteq \Bbb R^2$ iff it is of the form $G\cap S$ for some open set $G$ in $\Bbb R^2$.

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