I have to show that
is a topology. I know that this is not a topology, because it forms the basis for a topology on . But why is this the case? I am thinking of the case, for instance, . I don't think this is in , or am I incorrect?
Thanks in advance,
HTale.
OK, and also we have the analagous statement for unions, where we take the maximum. But, how about the empty set? You will never be able to get the empty set on this topology. The whole of the real line, yes but
Also, can I form the whole of the real line if I changed that to ?
Coming back to the original question, a set is said to be a base (or basis) for a topology if the collection of (arbitrary) unions of members of is a topology. The necessary and sufficient condition for a collection of subsets of a space to form a base for a topology is that the union of all the members of should be the whole space and that the intersection of any finite number of sets in should also be in .
In the case of the set above, it's easy enough to see that any finite intersection of half-open intervals is again a half-open interval. So is the base for a topology (the half-open interval topology).
Check out Chapter 2 of Part V of Topology and the Language of Mathematics (available for free here: Bobo Strategy - Topology). It gives some examples which you might find useful.