Difficulty understanding equivalent statement of Erdős Discrepancy Problem Recently I watched a famous youtube video of talk given by Terry Tao on Erdős Discrepancy Problem https://www.youtube.com/watch?v=QauoO0j9Y9Y.
I never heard of this problem before his announcement of proof, quite a beginner, right?
In the beginning of this talk, he gave an equivalent statement of the problem, which I can't really think out why, namely
Given an infinite sequence $f(1),f(2),\dotsc\in\{-1,1\}$ and $C>0$, there are $n$, $d$ such that
$\left|{\sum_{j=1}^nf(jd)}\right|>C$
the above statement is equivalent to
Given $C>0$, there is $N$ such that for every finite sequence $f(1),f(2),\dotsc,f(N)\in\{-1,1\}$ there are $n$, $d$ with $nd\leq N$ such that
$\left|{\sum_{j=1}^nf(jd)}\right|>C$
What flummoxes me is that how the above one implies the below one. If the above one is true (once conjecture but now proven theorem), does there have some properties telling us that it won't be too large the first $n$, $d$ (smallest $nd$) for every chosen sequence $f(1),f(2),\dotsc$ under a given $C$?
Perhaps this is obvious to most people. Or maybe my understanding of the statement is wrong. Hope someone will help out. Thanks.
 A: For a sequence $S = \{f(n)\}_{n \in \mathbb{N}}$, we call
$$
\max_{j, d, M} \left \lvert \sum_{j = 1}^M f(jd) \right \rvert
$$
the Discrepancy of the sequence $S$. So the first statement says that the discrepancy of any sequence of $\pm 1$ terms is unbounded. The second statement says that that for any $C$, there is a corresponding number $N = N(C)$ such that any sequence of at least $N$ many $\pm 1$ terms has discrepancy at least $C$.
It's clear that the second implies the first. It only remains to show that the first implies the second. I find it easier to show that if there are arbitrarily long, finite sequences with discrepancy at most $C$, then there is an infinite sequence with discrepancy at most $C$.
Then one uses the classic diagonal/compactness argument. Take a sequence of finite sequences with discrepancy at most $C$. Then there is a subsequence that converges pointwise. The resulting infinite sequence has discrepancy at most $C$.
This shows that the two formulations are equivalent.
I should add that this diagonal/compactness argument is very old and has very many names. It is in essence the Bolzano-Weierstrass Theorem, or perhaps Tychonoff's Theorem applied to $\{-1, 1\}^\mathbb{N}$. The theme is that $\{-1, 1\}^\mathbb{N}$ is sequentially compact, so it makes sense to pass to the limiting subsequence.
