I've been going through Munkres' Topology on my own, and I've come across an exercise where I can't even understand the question. It is exercise 7 of section 21 (p.134):
Let $X$ be a set, and let $f_n: X\rightarrow\mathbb{R}$ be a sequence of functions. Let $\bar{\rho}$ be the uniform metric on the space $\mathbb{R}^X$. Show that the sequence $(f_n)$ converges uniformly to the function $f:X\rightarrow \mathbb{R}$ if and only if the sequence $(f_n)$ converges to $f$ as elements of the metric space $(\mathbb{R}^X,\bar{\rho})$.
As far as I can tell, $\mathbb{R}^X$ is the set of all functions mapping $X$ to $\mathbb{R}$. I'm not exactly sure what is meant by the uniform metric on this set.
I suppose this is analogous to the way Munkres defined a $J$-tuple, when he defined infinite products, but to be honest I didn't fully understand that either. The uniform metric on $\mathbb{R}^X$ is like applying the uniform metric to a sequence of elements of $\mathbb{R}$ that is as long as the cardinality of $X$? But I completely don't understand the final phrase, 'the sequence $(f_n)$ converges to $f$ as elements of the metric space $(\mathbb{R}^X,\bar{\rho})$, I just have no guesses at what that could mean.
Best Answer
I apologize for my bad English. You can see the problem from two points of view. On the one hand, the phrase "...the sequence $(fn)$ converges uniformly to the function $f:X→R$ ..." means that there is a sequence of functions $f_n(x)$ such that $d(f_n(x), f(x)) < \epsilon$ for all $x$ and all $n>N$, where $N$ does not depend on $x$ (The classical definition of uniform convergence.) On the other hand, the phrase "...the sequence $(fn)$ converges to $f$ as elements of the metric space $(R^X,\overline ρ)$..." means that each element of the sequence $(fn)$ is a point of the metric space $(R^X, \overline ρ)$, and these points converge to $f$, which is also a point of $(R^X, \overline ρ)$. I hope this clarifies the question.