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Note: By upper/lower density, we shall mean the upper/lower asymptotic density as given here.

Question:

For any subset $S \subset \mathbb N$ with positive upper density, does there exists a $\varepsilon > 0$ such that the set

$$A_{\varepsilon} := \{z \in \mathbb N \ | \ \text {The set } S_z=\{n~|~nz \in S\}\text { has upper density at least } \varepsilon\}$$

has nonzero lower density?

Remarks:

1. This question arose when trying to prove a number theory result involving greatest common denominators of sets of naturals.

2. The corresponding result, if one replaces the set $A_\varepsilon$ with the set $B := \{z \in \mathbb N \ | \ nz \in \text {for infinitely many } n \in \mathbb N \}$ is true, and not overly difficult to prove. In fact we even have an explicit bound for the lower density of $B$ - it is at least the upper density of $S$.

Note: By upper/lower density, we shall mean the upper/lower asymptotic density as given here.

Question:

For any subset $S \subset \mathbb N$ with positive upper density, does there exists a $\varepsilon > 0$ such that the set

$$A_{\varepsilon} := \{z \in \mathbb N \ | \ \text {The set } S_z=\{n~|~nz \in S\}\text { has upper density at least } \varepsilon\}$$

has nonzero lower density?

Remarks:

1. This question arose when trying to prove a number theory result involving greatest common denominators of sets of naturals.

2. The corresponding result, if one replaces the set $A_\varepsilon$ with the set $B := \{z \in \mathbb N \ | \ nz \in S \text { for infinitely many } n \in \mathbb N \}$ is true, and not overly difficult to prove. In fact we even have an explicit bound for the lower density of $B$ - it is at least the upper density of $S$.

Note: By upper/lower density, we shall mean the upper/lower asymptotic density as given here.

Question:

For any subset $S \subset \mathbb N$ with positive upper density, does there exists a $\varepsilon > 0$ such that the set

$$A_{\varepsilon} := \{z \in \mathbb N \ | \ \text {The set of } n \in \mathbb N \text { such that } nz \in S\text { has upper density at least } \varepsilon\}$$

has nonzero lower density?

Remarks:

1. This question arose when trying to prove a number theory result involving greatest common denominators of sets of naturals.

2. The corresponding result, if one replaces the set $A_\varepsilon$ with the set $B := \{z \in \mathbb N \ | \ nz \in \text {for infinitely many } n \in \mathbb N \}$ is true, and not overly difficult to prove. In fact we even have an explicit bound for the lower density of $B$ - it is at least the upper density of $S$.

Note: By upper/lower density, we shall mean the upper/lower asymptotic density as given here.

Question:

For any subset $S \subset \mathbb N$ with positive upper density, does there exists a $\varepsilon > 0$ such that the set

$$A_{\varepsilon} := \{z \in \mathbb N \ | \ \text {The set } S_z=\{n~|~nz \in S\}\text { has upper density at least } \varepsilon\}$$

has nonzero lower density?

Remarks:

1. This question arose when trying to prove a number theory result involving greatest common denominators of sets of naturals.

2. The corresponding result, if one replaces the set $A_\varepsilon$ with the set $B := \{z \in \mathbb N \ | \ nz \in \text {for infinitely many } n \in \mathbb N \}$ is true, and not overly difficult to prove. In fact we even have an explicit bound for the lower density of $B$ - it is at least the upper density of $S$.

Note: By upper/lower density, we shall mean the upper/lower asymptotic density as given here.

Question:

For any subset $S \subset \mathbb N$ with positive upper density, does there exists a $\varepsilon > 0$ such that the set

$$A_{\varepsilon} := \{z \in \mathbb N \ | \ \text {The set of } n \in \mathbb N \text { such that } nz \in S\text { has upper density at least } \varepsilon\}$$

has nonzero lower density?

Remark: This question arose when trying to prove a number theory result involving greatest common denominators of sets of naturals.

Note: By upper/lower density, we shall mean the upper/lower asymptotic density as given here.

Question:

For any subset $S \subset \mathbb N$ with positive upper density, does there exists a $\varepsilon > 0$ such that the set

$$A_{\varepsilon} := \{z \in \mathbb N \ | \ \text {The set of } n \in \mathbb N \text { such that } nz \in S\text { has upper density at least } \varepsilon\}$$

has nonzero lower density?

Remarks:

1. This question arose when trying to prove a number theory result involving greatest common denominators of sets of naturals.

2. The corresponding result, if one replaces the set $A_\varepsilon$ with the set $B := \{z \in \mathbb N \ | \ nz \in \text {for infinitely many } n \in \mathbb N \}$ is true, and not overly difficult to prove. In fact we even have an explicit bound for the lower density of $B$ - it is at least the upper density of $S$.