2006-07-20 05:08:42 · 10 answers · asked by SUVAJIT 1 in Science & Mathematics ➔ Mathematics
The null set is a set of measure zero, this is not necessarily the empty set. So if the null set is in a metric space, it could consist of a set of points. If this set is a finite set of points then this null set is closed. If you mean the null set to be the empty set or the set with nothing in it, then the null set is both open and closed.
http://en.wikipedia.org/wiki/Null_set
2006-07-20 05:23:15 · answer #1 · answered by raz 5 · 0⤊ 0⤋
Yes, Eulercrosser gave the correct answer.... but to add something I will say that the null set is also closed by definition. This is because the compliment of an open set is always closed and since the whole space is in the topology, the whole space is also open, and so the null set (the complement of the whole space) is closed.
You can further reason that the whole space is also always open and closed.
ahh, yeah... I was thinking in terms of the empty set in a topology also. Raz has a point.
2006-07-20 12:25:17 · answer #2 · answered by random.oracle_23 2 · 0⤊ 0⤋
If you have defined 'open set' with the topological definition, then the previous posters are right, the empty set is open by definition in a topology.
If you have defined 'open set' from the metric space definition, then a set U is open if for all x in U there is an ball around x contained entirely in U. If U is empty, this is true because there are no x in U, so it is an empty proposition.
Basically, if U is empty, then the statement "For all x in U, P(x)" is true for all propositions, P.
2006-07-20 12:32:37 · answer #3 · answered by thomasoa 5 · 0⤊ 0⤋
Yes
2006-07-20 12:16:41 · answer #4 · answered by hackmaster_sk 3 · 0⤊ 0⤋
Yes, by definition of a topology.
Definition of a topology:
Let U be a collection of subsets of X called "open." Then U is a topology if:
1) Ã, X are in U
2) For every U1, U2, U3, . . . (could be infinite) that are in U, the union of these is in U.
3) For every U1, U1, U3, . . ., Un (finite) that are in U, the intersection of these is in U.
By part 1), à is in U, so à is open.
2006-07-20 12:10:55 · answer #5 · answered by Eulercrosser 4 · 0⤊ 0⤋
I think it depends on the phrasing of both of these set concepts. My first impression would be yes.
2006-07-20 12:11:58 · answer #6 · answered by molex77 3 · 0⤊ 0⤋
yes
2006-07-20 12:15:12 · answer #7 · answered by wooddale 2 · 0⤊ 0⤋
yes
2006-07-20 12:11:26 · answer #8 · answered by raj 7 · 0⤊ 0⤋
It depends if rational or imaginary
2006-07-20 12:14:58 · answer #9 · answered by Ron K 3 · 0⤊ 0⤋
Yeah, same thing.
2006-07-20 12:11:15 · answer #10 · answered by Pirate_Wench 5 · 0⤊ 0⤋