I am studying Topology and Groupoids by Ronald Brown, 3rd Edition. I believe I have spotted an errata in a given question, as I seem to have found a counterexample to what it wishes me to prove. I wish to know if I am correct or have misunderstood.
In the book, the following appears as exercise 3.3.8 (I have broken the question into parts of ease of reading):
Let $A$ be a nonempty subset of a metric space $X$. We say $A$ is bounded if $sup\{d(x, y) : x,y\in A\}$ exists and then this number is called the diameter of $A$.
Let $x, y\in X$. If there is a connected set containing $x, y$ of diameter $<1$, let $\sigma(x, y)$ be the infinum of the diameter of such sets. If no such set exists, let $\sigma(x, y) = 1$.
(a) Prove that $(x, y)\mapsto \sigma(x, y)$ is a metric on $X$.
(b) Let $L\subseteq X$ be the subset of $X$ of points at which $X$ is locally connected. Prove that $\sigma$ induces the discrete topology on $X\setminus L$ (if $X\setminus L \neq \emptyset$) and that $\sigma$ induces the same topology on $L$ as does $d$.
(c) Prove that in $L$ the open ball for the metric $\sigma$ is connected if of radius $<1$.
I have proven (a), my problem is with (b). If I have not found an error, I think understanding why my counterexample does not work would be instructive:
Let $X$ be the "rational hair-comb" $(\mathbb{R}\times\{0\})\cup \bigcup_{z\in \mathbb{Q}} \{z\} \times [0, 1]$. This set is connected, but outside of the $x$ axis $X$ is not locally connected as any sufficiently small open ball away from the $x$ axis will intersect with countably many disconnected vertical lines. We then have $L = \mathbb{R}\times\{0\}$.
The question seems to indicate therefore that $X\setminus L$ together with $\sigma$ should have the discrete topology, but as far as I can tell for any $z_1 = (a, b), z_2 = (a, c)\in X$ on the same vertical line we will have $\sigma(z_1, z_2) = d(z_1, z_2) = |b - c|$ so long as $|b - c| < 1$.
Therefore every open ball of radius $\epsilon$ under $\sigma$ around $z_1$ will contain points distinct from $z_1$, so that $X\setminus L$ cannot have the discrete topology.
Is this correct, or is there something I am missing with this question?
Thanks for the feedback from Nethesis and Lee Mosher.
After thinking about this for a while, it seems like your counterexample is valid. Perhaps one must require that $x$ and $y$ are contained in a connected neighborhood, rather than just a connected subset.