Is the empty set a compatible subset of a dense set?

[HallsofIvy]

Simply that if every subset of T (as 'morphism' says in post 24) is dense in itself, then the empty set must be dense in itself as a subset of T. Since the empty set contains no points of the point set T, this would appear to me to be a contradiction.

No, the usual definition of a dense set is: "If set X is dense in set Y, any point in set Y can be 'well-approximated' by a point in set X". If set Y is the empty set, that is trivially true, since there are no points in Y to worry about.

Or, a little more generally, a set X is dense in set Y if and only if Y is contained in the closure of Y. As has been said repeatedly, every set, no matter what it is a subset of, is dense in itself because every set is contained in its own closure. The empty set is itself a closed set: The closure of the empty set is itself which certainly contains itself. That has nothing at all to do with whether it contains any points.

 

[SW VandeCarr]

Or, a little more generally, a set X is dense in set Y if and only if Y is contained in the closure of Y. As has been said repeatedly, every set, no matter what it is a subset of, is dense in itself because every set is contained in its own closure. The empty set is itself a closed set: The closure of the empty set is itself which certainly contains itself. That has nothing at all to do with whether it contains any points.

Thank you Quadaphonics and HallsofIvy. The key, at least to me, is that the empty set is in fact considered 'dense in itself.'

 

[HallsofIvy]

Since the definition of "closure" of a set is the set itself union all limit points, every set is contained in its own closure: every set is dense in itself. I don't see how that is "key" to anything!

The whole point of "density", typically, is to have some comparatively small set dense in a large set (as the countable rationals in the uncountable reals).