Bishop & Goldberg say exactly what I did, they just add the example of an open neighborhood.
Quote:
Originally Posted by Ben
I will happily concede my original definition was wrong iff it can be shown that, for any pair, say, of closed sets  that  . I doubt it, but I am not completely sure. Someone help me out here! |
That clearly can't be, but wouldn't be necessary for stating that a neighborhood needn't be open.
Any set comprising a neighborhood of a point (as a subset) is also a neighborhood of the same point.
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Inutil insegnŕ al mus, si piart timp, in plui si infastiděs la bestie.
Hypography Forum PITA...... er, Administrator.
