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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.

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    $\begingroup$ The uniform metric $\bar p$ is defined by $\bar p(f,g)=\sup_{x\in X}(\min(|f(x)-g(x)|,1)$ $\endgroup$ Sep 1, 2015 at 8:34
  • $\begingroup$ @StefanHamcke I think you can post that as an answer. $\endgroup$ Sep 4, 2015 at 12:41
  • $\begingroup$ Read the definition on p.124 of Munkres's Topology. (You may find it useful to first check the index if you come across a term you are unfamiliar with when reading a text.) $\endgroup$
    – user642796
    Sep 4, 2015 at 12:43
  • $\begingroup$ Alright, that helps, but I still don't fully understand what he means by 'the sequence converges to $f$ as elements of the metric space.' What does it mean to converge as an element? $\endgroup$ Sep 4, 2015 at 17:03
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    $\begingroup$ It means that $d(f_n, f) \to 0$ as $n\to \infty$, where $d$ is the metric. $\endgroup$
    – user169852
    Sep 4, 2015 at 17:11

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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.

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