non metrizability of pointwise convergence I have some question concerning the topology of pointwise convergence, when I was trying to understand the following paper titled:  A NOTE ON POINTWISE CONVERGENCE by Marion K Fort, Jr:
http://www.ams.org/journals/proc/1951-002-01/S0002-9939-1951-0040647-7/S0002-9939-1951-0040647-7.pdf
The paper tries to show that there does not exists a metric $d$ for $C$ with the property that $lim_{m,n \rightarrow \infty} d(f_{n},f_{0}) = 0$ if and only if the sequence $f_{n}$ converges pointwise to $f_{0}$.
I understand this question has been asked before as well as this paper has been referred to in the answers on this forum.  I came across this question a long time ago in an homework assignment and later i tried to read and understand the paper on and off but have not succeed due to the way the paper was written.
I was wondering if I can get some help with demystifying some of my confusion.
For each point I need help with, i have referenced the paper with paragraph number.  
On paragraph 6, it states:  "We now let $T$ be the collection of all intervals $J$ such that for some $n$, $J$ is an interval of the subdivision $S_{n}$."  Here if I have a set $T$, do I pick one interval $J$ for every number $n$, also how is the interval $J$ chosen?
On paragraph 7, "Let $n$ and $m$ be positive integers.  We define the function $f_{{n},{m}}$.  If $J$ is an interval of the subdivision $S_{n}$, we define $\epsilon_{J} = 1$ if $k(J) \leq m$ and $\epsilon_{J} = 1/k(J) > m$.  Let $f_{{n},{m}}$ be the function on $[0,1]$ which is of type ($m$, $\epsilon_{J}$) on each interval $J$ of $S_{n}$.  The graph of $f_{{n},{m}}$ has a hump on each interval of $S_{n}$.  It is easily seen that $f_{{n},{m}}$ is continuous, however, since if $\delta > 0$ then the graph of $f_{{n},{m}}$ can contain at most a finite number of humps of height greater than $\delta$.
My question are as follows, from this point on in the paper, are $m$, and $n$ fixed positive integers?
The way the function $f_{{n},{m}}$ is defined, is there suppose to some kind of analytic expression for it, because I am seeing a function $k:T \rightarrow N$ where $k(J)=n$, and also the function $\epsilon_{J}$.  I am guessing $f_{{n},{m}}$ is defined as $f_{{n},{m}}(\epsilon_{J})$ which I am not sure if it is correct.  Also, how is the function being type ($m$, $\epsilon_{J}$) related to how $f$ is defined earlier being of type ($n$, $\epsilon$).
For paragraph 8, I am not sure how the author conclude that $f_{{n},{m}} = 0$ for all but at most three values of $m$.
for paragraph 9, I don't understand how the value for $N_{n}$ is determined.
Thank you in advance for any help you can give me.  
 A: The paper is written in the terse style preferred by journals and professional mathematicians (and to be honest, I quite like it as well even though it requires that the reader keeps stopping and checking that they've really understood what was just said).  It gets easier to read with practice, and asking questions about the material in here is expected: don't get disheartened.
Answering the questions as you've indexed them:


*

*paragraph 6: this is actually a statement about the structure of $T$, not the numbers $n$.  The elements of $T$ are intervals, referred to generally as $J$, and if you pick any such $J$ then it must belong to an $S_n$ for some $n$, no matter what that $n$ is.  What matters here is that all the elements of $T$ are like this: you're not choosing $T$, it's being defined for you.  If you wanted to choose a specific one (perhaps to work an example through) then you would pick a value for $n$ and then look at $S_n$ and then choose endpoints for $J$ such that $J \subseteq S_n$.  This is then an element of $T$ by construction.

*paragraph 7: $n$ and $m$ are positive integers, yes.  They are fixed in the sense that whenever $f_{n,m}$ is referred to you could replace them with fixed positive integers, but the exact choice may vary from one usage of $f_{n,m}$ to the next (e.g. one example might work easily with $f_{1,3}$ but the next might work better with $f_{100,300}$, especially since we're thinking about convergence).

*para7 cont'd: $f$ is analytic but it would be hard to write down an explicit expression for it, which is why it's being defined in this way.  In analysis we often find pathological functions (those that behave badly enough to act as counter-examples ;-) ) that oscillate wildly or jump erratically and it's easier to describe them than to write them out in symbols.

*para7 cont'd: $f_{n,m}$ is not $f_{n,m}(e_J)$; rather $f_{n,m}$ is given by the original function $f$ applied to values $m$ and $e_J$.  That's what's meant by saying that $f_{n,m}$ is of type $(m,e_J)$.  By the definition of a function of type $(n,\epsilon)$ we therefore have that:


*

*the domain of $f_{n,m}$ contains $[a,b]$

*the graph of $f_{n,m}$ restricted to $[a,b]$ consists of the piecewise continuous line joining $(a,0)$, $(x_m, 0)$, $(x_{m+1},e_J)$, $(x_{m+2},e_J)$, $(x_{m+3},0)$ and $(b,0)$ where $e_J = 1$ if $k(J) \leq m$ and $e_J = 1/k(J)$ if $k(J) > m$.  Thus each $f_{n,m}$ is a hump function whose exact shape depends on the injective function $k:T\rightarrow {\mathbb Z}^+$


*paragraph 8: we fix $n$ first, so we have restricted ourselves to those $J$ that live in $S_n$ for this fixed $n$.  That determines the values that $m$ can take (because $S_n$ is a normal subdivision that defines exactly what values $m$ can take on), and so this restricts $f(t)$ for $0\leq t\leq 1$ to just three non-zero values.  For any fixed $n$ then, $f_{n,m}=0$ for all but at most three values of $m$.

*paragraph 9: $f_0$ is defined to be identically zero so that $\lim_{n\rightarrow \infty} d(f_{n,m},f_0) = 0$ for each positive integer $n$.  This means that the $f_{n,m}$ are converging to $f_0$ so when $m$ is large enough we know that $f_{n,m}$ and $f_0$ are as close together as we like. Since $n$ is fixed we can do this for each value of $n$.  So, pick $N_n$ so that $d(f_{n,m},f_0) < 1/n$ (for each $n$) (for any given example function $f_{n,m}$ you write down you can find the right value of $N_n$).
You're still going to need to think about all of this before it really makes sense, I suspect.  Try picking a value $n$ (say $n=4$) and an interval $J$ in $S_n$ and a function $k$ and draw out the graph of $f_{n,m}$ -- that picture should help put all these words into context.
EDIT: your comment says you're still confused a little, so:
\begin{eqnarray}
S_1 & = & \{0,1\}\quad \mbox{by hypothesis,} \\
S_2 & = & \left\{0,\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \ldots, 1\right\} \\
S_3 & = & \left\{0, \frac{1}{4}, \frac{3}{8}, \ldots, \frac{1}{2}, \frac{5}{8}, \frac{11}{16}, \ldots, \frac{3}{4}, \ldots, 1 \right\} \\
S_4 & = & \left\{0, \frac{1}{8}, \frac{3}{16}, \ldots, \frac{1}{4}, \frac{5}{16}, \frac{11}{32}, \ldots, \frac{3}{8}, \ldots, 1\right\}
\end{eqnarray}
Note that each $S_n$ refines the previous $S_n$ by iterating the subdivision procedure over each new interval formed.
The $f_{n,m}$ are then defined on the $S_n$, and here is a picture of $f_{3,1}, f_{3,2}$ and $f_{3,3}$ where I assume that $k(J) = 2$ for convenience:

It should now be clear (look at the graph between about $0.23$ and $0.25$) that the functions overlap at most three-at-a-time, which is why $f_{n,m}$ is non-zero for at most three values of $m$.  You might want to add in $f_{3,4}$ and $f_{3,5}$ yourself to convince yourself of that.
