The Question: Consider the Metric Space $\chi = ${($x_1,x_2,...,x_N$)}, with the word metric d(x,y) = number of digits that differ between x and y. Are the following sets open, closed, or neither?
(i) $S_1$ = {x$\in\chi$|$x_i=0,i=1,...,5$}
(ii) $S_2$ = {x$\in\chi$|$x_{i+1} \neq x_i,i=1,...,N-1$}
My attempt:
Take x $\in S_1$. Then, taking r < 1, $B_r(x) = $ {x} $ \in S_1$ Thus, $S_1$ is open. But, for r < 1, $B_r(x)$, $B_r(x)/${x} = $\emptyset$. Thus, $S_1$ has no cluster points, and by extension contains all of its cluster points (none). Thus, $S_1$ is also closed.
Similarly, I get that $S_2$ is both open and closed. (I also used the argument that $S_2$ is finite to show it was closed, since it should only have two elements - $x_1$ = 101010... and $x_2$ 010101...)
Is my logic flawed, or was it a problem with how the question was worded (that both open and closed is not an option)? Also, more generally (not part of the problem), but is every subset of this space $\chi$ both open and closed?
There's nothing wrong with your logic. In each underlying space, you've demonstrated that every singleton is open, which means that every subset of each underlying space is a union of open sets. So every possible subset of each component is open (so every possible subset of each component is also closed), and the same holds true in the product topology.
In short, this set is literally as disconnected (defined as a topological space that can be divided into two disjoint open components) as you can get.