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Math-5532

Erdős Problem #255

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Statement

Given any sequence $z_1,z_2,\ldots\in[0,1]$, must some fixed interval $I$ satisfy \[ \limsup_{N\to\infty} \left|\#\{n\leq N:z_n\in I\}-N|I|\right|=\infty? \] Yes. We give the localization argument that passes from a uniform discrepancy lower bound to a single fixed interval. The established uniform lower bound is an explicit external theorem; the quantifier change is proved below.

Restatement from Math-5532-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

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Math-5532-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5532

Erdős Problem 255: one fixed interval with unbounded discrepancy

1) Statement and scope

Given any sequence $z_1,z_2,\ldots\in[0,1]$, must some fixed interval $I$ satisfy \[ \limsup_{N\to\infty} \left|\#\{n\leq N:z_n\in I\}-N|I|\right|=\infty? \] Yes. We give the localization argument that passes from a uniform discrepancy lower bound to a single fixed interval. The established uniform lower bound is an explicit external theorem; the quantifier change is proved below.

2) The uniform discrepancy input

We use the following weaker consequence of Schmidt's discrepancy theorem: for every infinite sequence $y_j\in[0,1)$, \[ \sup_{N\geq1,\ 0\leq t\leq1} \left|\#\{j\leq N:y_j<t\}-Nt\right|=\infty. \] Schmidt's 1972 Theorem 1 supplies a quantitative logarithmic lower bound and hence this assertion. By itself, a statement with the supremum over $t$ does not select a fixed interval.

3) Baire localization

Suppose, to obtain a contradiction, that every anchored interval $[0,x)$ has bounded discrepancy as $N$ varies. Put \[ D_N(x)=\#\{n\leq N:z_n<x\}-Nx,\qquad X=(0,1)\setminus\{z_n:n\geq1\}. \] The space $X$ is a dense $G_\delta$ subset of $(0,1)$ and is a Baire space. For each $N$, the function $D_N$ is continuous on $X$. For an integer $m\geq1$, let \[ E_m=\{x\in X:|D_N(x)|\leq m\text{ for all }N\}. \] These sets are closed in $X$ and cover $X$ by the supposition. The Baire category theorem implies that some $E_m$ contains $J\cap X$, for a nonempty open interval $J\subset(0,1)$.

For fixed $N$ and $x\in J$, approach $x$ from the left by points of $X\cap J$. The count with the strict inequality $z_n<x$ is left-continuous. Passing to the limit gives $|D_N(x)|\leq m$. This holds for every $N$ and every $x\in J$, including points repeated in the sequence. Removing the countable jump set before applying Baire avoids a continuity assumption at those points.

4) Extracting and rescaling the visits

Choose $a<b$ in $J\cap X$, and put $\ell=b-a$. The number of visits to $[a,b)$ up to time $N$ is \[ N\ell+D_N(b)-D_N(a)=N\ell+O(m), \] so there are infinitely many such visits. Let their indices be $n_1<n_2<\cdots$, and define $y_j=(z_{n_j}-a)/\ell\in[0,1)$. At time $n_j$, there have been exactly $j$ visits. For $0\leq t\leq1$, setting $x=a+t\ell$ gives the exact identity \[ \#\{i\leq j:y_i<t\}-jt =D_{n_j}(x)-(1-t)D_{n_j}(a)-tD_{n_j}(b). \] Every point on the right lies in $J$, so the absolute value is at most $2m$, uniformly in $j,t$. This contradicts the external uniform discrepancy theorem. Therefore some fixed $x$ has unbounded $|D_N(x)|$. An unbounded sequence of these values has limsup infinity, because each finite initial segment is bounded. The interval $I=[0,x)$ answers the question.

5) Sources and scope

Sources: https://www.erdosproblems.com/255, which records Schmidt's original 1968 fixed-interval result; W. M. Schmidt, Irregularities of distribution, VII, Acta Arith. 21 $1972$, 45--50, Theorem 1, https://doi.org/10.4064/aa-21-1-45-50. The external inputs are Schmidt's uniform lower bound and the standard Baire category theorem. The reduction above establishes the full fixed-interval assertion from them, without claiming a new resolution or local formal verification.

6) Completion estimate

COMPLETION: 100%

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