Discussion
Let X be a metric space and \(Y \subset X\) :
Answer: B
The true statement is if Y is nowhere dense in X, then \(X \backslash Y\) is dense in X (the paper prints 'nowhere in X').
If Y is nowhere dense, its closure \(\bar{Y}\) has empty interior, so the open set \(X \backslash \bar{Y}\) is dense. Since \(X \backslash Y \supseteq X \backslash \bar{Y}\), \(X \backslash Y\) is dense too.
The others fail with \(X = \mathbf{R}\), \(Y = \mathbf{Q}\): Y is dense, yet \(X \backslash Y\) (the irrationals) is dense, not nowhere dense; \(\mathrm{Int}(\mathbf{Q}) = \phi\); and \(\mathrm{Int}(\mathbf{R} \backslash \mathbf{Q}) = \phi\).
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