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I have two statements

"A set X is nowhere dense"

"A set X has no interior"

Are these both equivalent statement?

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    $\begingroup$ Not quite: a set is nowhere dense if its closure has empty interior. $\endgroup$ Aug 8, 2016 at 19:16
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    $\begingroup$ Note that having empty interior is not the same as having no interior. Every subset of a topological space has an interior (which may or may not be empty). The empty set (in the mainstream mathematics) does exist. $\endgroup$
    – tomasz
    Aug 8, 2016 at 19:28
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    $\begingroup$ To add to @BrianM.Scott 's comment, the two notions are equivalent for closed sets. $\endgroup$ Aug 9, 2016 at 0:08

2 Answers 2

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No, $\mathbb{Q}\subset\mathbb{R}$ has an empty interior, but is dense in $\mathbb{R}$.

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For possible future use by others, it might be helpful to dissect these notions a bit more systematically in a metric space setting to see what makes them different. In particular, this will show how being nowhere dense is a super-strong way of having empty interior.

$B(x,\epsilon)$ denotes the open ball centered at $x$ with radius $\epsilon,$ where $\epsilon > 0$ is assumed.

$X$ has empty interior

This is equivalent to each of the following:

$X$ has no interior points.

For each $x$ in the space, $x$ is not an interior point of $X.$

For each $x$ in the space, there is no nonempty open set containing $x$ that is a subset of $X.$

For each $x$ in the space, there is no ball $B(x,\epsilon)$ that is a subset of $X.$

For each $x$ in the space, each ball $B(x,\epsilon)$ contains at least one point that belongs to the complement of $X.$

$X$ is nowhere dense

This is equivalent to each of the following:

$X$ is not dense in each nonempty open set.

$X$ is not dense in each ball $B(x,\epsilon).$

For each $x$ in the space, each ball $B(x,\epsilon)$ is NOT densely filled with points from $X.$

For each $x$ in the space, each ball $B(x,\epsilon)$ contains at least one nonempty open set that is a subset of the complement of $X.$

For each $x$ in the space, each ball $B(x,\epsilon)$ contains at least one ball all of whose points belong to the complement of $X.$

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