Consider the following conjecture of mine:
If $Y$ is a subspace of $X$ whose subspace topology is discrete, then $Y$ is closed in $X$.
If I'm not mistaken, $A = \{\frac{1}{n} \mid n \in \Bbb{N}$ is a counterexample. Since $\{1\} = A \cap (\frac{3}{4},2)$ and $\{\frac{1}{n}\} = A \cap (\frac{1}{n+1},\frac{1}{n-1})$ for all $n \ge 2$, all singletons in $A$ are open so $A$'s subspace topology is discrete. But $A$ is not closed in $\Bbb{R}$.
But this is rather confusing. Since all discrete (metric) spaces are complete, and since a subspace of a complete metric space is closed iff it is complete, shouldn't $A$ actually be closed? What have I done wrong?
Best Answer
The discrete metric refers to a particular metric on a space, that where $d(x,y) = 1$ for $x \neq y$. While the metric on your subspace generates the same discrete topology, it is not the same as the discrete metric and therefore doesn't need to be complete. Completeness is only a property of the metric, not the topology.
(If you look at the question you linked to, you'll notice that the key step is that if $d(x_n,x_m)< \frac{1}{2}$ then $x_n=x_m$. This is true for the discrete metric, but not true for your set.)