Prove that (0,1] is open in [-1,1] but not in R?
The point 1 does not have a neighbourhood in R that does not include elements that are not in (0,1].
Too many negatives. Try again.
All neighbourhoods of 1 in R contain elements that are not in (0,1]. However, in [-1,1] this is not the case as there are no elements of [-1,1] that are not in a neighbourhood of 1.
At the other end of the interval (0,1], the end is open. Every point near 0 has a neighbourhood completely contained in (0,1].
So every point of (0,1] has a neighbourhood entirely within (0,1] except 1, for which see above.
I guess that's the intersection.. he means of course that if you use the subspace topology then a set is open if it is the intersection of he subset and an open set in the base set.
eg (0,1] is open in [1,1] because (0,2) is open in R, and (0,2) intersected with [0,1] = (0,1]..