Meagre subsetes of [0,1]

A subset $A$ of a topological space $X$ is nowhere dense if it's closure has empty interior. It is true that every closed nowhere dense set is nowhere dense. A set is meagre if it is a countable union of nowhere dense sets.

What is an example of a closed meagre subset of $[0,1]$ which fails to be nowhere dense?

• Explain the downvotes please so I know what is wrong with the question. If it is too obvious, please have the decency to post something so I know. – J.K.T. May 20 '15 at 19:38
• It's too broad. A good answer depends on what else you know about the set. As is, about all that can be said is that it's nowhere dense if it satisfies the definition of nowhere dense. That is, out of context, knowing that the set is meager does you no good for showing that it is nowhere dense. In a specific example, you could perhaps show that the countable cover that you used to prove that the set is meager can actually be taken to be a finite cover, in which case the set will actually be nowhere dense. But there is no guarantee that this approach is always useful. – Ian May 20 '15 at 19:40
• @Ian Very well. Then is there a closed meagre set which fails to be nowhere dense? If my initial question was too broad the revised one is bound to be more tractable and would satisfy my curiosity. – J.K.T. May 20 '15 at 19:47
• It actually cannot be closed; a closed meager set has empty interior, so it is nowhere dense. – Ian May 20 '15 at 19:47
• @Ian Then it seems my original question was not too broad. I am interested only in closed sets. – J.K.T. May 20 '15 at 19:50

Consider the three properties: closed, meager, not nowhere dense. There are sets with any two of these properties but no sets with all three. Examples:

• Closed+meager: A finite set; a Cantor set.
• Closed+not nowhere dense: a closed interval.
• Meager+not nowhere dense: The rationals; a union of fat Cantor sets with full measure.

We cannot have all three properties, because:

1. If a set is meager, then by the Baire category theorem it has empty interior.
2. If a set is not nowhere dense, then its closure has nonempty interior.
3. If a set is closed, then it is equal to its closure.

So having all three properties gives a contradiction (provided the underlying space is a complete metric space).