Show that ${\mathscr C}(\{1,..,n\},R)$ and $R^n$ have the same open sets Question:
Let X be the set $\{1,2,...,n\}$ equipped with the discrete metric ($\delta(x,y)=0$ if $x=y$, $\delta(x,y)=1$ if $x\neq y$). Then ${\mathscr C}(X, R)$ and $R^n$, where $R$ is the real numbers, are two Banach spaces which are essentially identical as real linear spaces but which have different norms. Show that they have the same open sets.
My answer:
${\mathscr C}(X, R)$ is the space of functions defined on  $\{1,2,...,n\}$ with the discrete metric so $f(1), f(2), ..., f(n)$ gives an n-tuple of real numbers. Under the discrete metric, any function is continuous.
Let A be an open set in $R^n$, then if A is empty then A is also an open set in ${\mathscr C}(X, R)$. If A is not empty, let $a\in A$ then there is $r>0$ such that $S_r(a)\subseteq A$. If $r<1$ then let $(a_1, a_2, ..., a_n)$ be element of X then $S_r(a)=\{(a_1, a_2, ..., a_n)\}$ which is an open set, since all sets are open under the discrete metric. If $r\geq 1$ then $S_r(a)=X$ since all points in X are distance $\leq 1$ from a. Since X is an open set, $S_r(a)$ is open in ${\mathscr C}(X, R)$.
Let A be an open set in ${\mathscr C}(X, R)$. As above if A is empty, A is open in $R^n$. If A is non-empty let $a\in A$ then there is $r>0$ such that $S_r(a)\subseteq X$.
My question:
Does the above argument make sense? Particularly the last paragraph, I am not sure what I am trying to prove.
[For reference, the question is Problem 15-2 of "Introduction to Topology and Modern Analysis" by G.F. Simmons.]
 A: You have consider the metric on ${\mathscr C}(X, R)$: the distance between two functions $f,g$ in it, is $\sup \{|f(x) - g(x)\mid x \in X \}$, which is well-defined when $X$ is compact (as then all continuous real-valued functions are bounded, so the supremum exists).
The topology on $X$ itself ($X$ needn't even be metric for this definition to work) only serves to determine which real-valued functions are continuous, the metric on ${\mathscr C}(X, R)$ is derived from that of $R$: we take the distances $|f(x) - g(x)|$ (in $R$) and take the largest of these. 
So your remark that all functions from $\{1,2,\ldots,n\}$ to $R$ are continuous, is correct, and this is the only place we use that the domain space $X$ is discrete. We can indeed identify them with $n$-tuples of reals, as we will do.
By the general definition, the distance in ${\mathscr C}(X, R)$ between $(a_1,\ldots, a_n)$ and $(b_1,\ldots,b_n)$ is just $\sup \{|a_i - b_i|: i = 1,\dots,n\}$. Draw what the open ball around a point (do $n=2$ for the intuition, for $n=1$ we see right away that both spaces are just $R$ in the usual metric) looks like.
Then show that any standard Euclidean ball around a point, contains a $\sup$-metric ball around that point and vice versa.
