VIP topology on $X$

Question

I came across the following: Consider the VIP topology on $X$:

$$\tau=\{U \subset X: U=\emptyset \;\;\text{or}\;\;a \in U \}$$ where $a \in X$ is fixed.

The author proves every non empty open set is dense in $X$. After he proves every proper subset is nowhere dense.

How the second one is true ? because take any proper non empty open set $U$, and using first one, $U$ is dense. But by the second one $U$ is nowhere dense. What's going on? Is I'm misunderstand something or it is a typo in a book?

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I add a reference to it: See 4 and 7

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Here's his proof:

Answer 1

The proof you post clearly shows that we assume that $\overline A$ is a proper subset of $X$, rather than just $A$ itself.

So yes, this is in fact a typo. In other words, the proof shows that if $A$ is not dense, then it is nowhere dense. Which is to be expected.

Answer 2

There are two types of subsets $A$ of $X$:

$a \in A$, then $A$ is open by definition, and dense as all other non-empty open subsets also contain $a$ and so intersect $A$.

$a \notin A$, then $A$ is closed (as $X\setminus A$ then does contain $a$ so is open) and contains no non-empty open set (or it would also have contained $a$), so $\operatorname{int}(\overline{A}) = \operatorname{int}(A) = \emptyset$ and $A$ is nowhere dense.

So not every proper subset is nowhere dense, just all the non-open ones.

Note that this set is a so-called "door space": every subset is open or closed (or both). This is quite a rare property, really.