- #1

kahwawashay1

- 96

- 0

^{n}

point a is in S: a[itex]\in[/itex]S

The point "a" is an interior point of S if there is an open n-ball with center "a", all of whose points belong to S.

ie., every interior point of S can be surrounded by an n-ball such that B(a)[itex]\subseteq[/itex]S, where B(a) is the set of all points x in R

^{n}such that ||x-a||<r (r is radius of the ball)

Ok so let's say we have such an S with point "a" and also containing all of the points in B(a).

Now, according to Apostol, the "interior" of S is the set of all interior points of S and is denoted by int S.

So let Q=int S for the S that I defined above. So, Q={a}, since "a" is the only interior point of S.

Now comes "open set"

A set W is an open set if all its points are interior points. ie., W is open set if and only if W = int W

This is what I don't understand. And actually, instead of "W", Apostol used S again, but I do not think he means any relation with the previous S's?

I know that an open set is basically an open interval, like (3, 4) on the x-axis would mean 3<x<4. So shouldn't an open set just be B(a)? Like in the case of S=(3, 4), S is an open set with interior point a=3.5, since 3.5 is the center, and B(a) is all x such that ||x-3.5||<0.5 ..but that means that S contains points other than just int S; namely, it also contains all points satisfying B(a), which are not interior points of S, so S≠int S

?