Let A and B be opensubset of R prove the union is open Solution Let O subset R b

Question

Let A and B be opensubset of R prove the union is open

Explanation / Answer

Let O subset R be open. (1) For x in O, let Rx=inf{ r>=0 : N(x,Rx) contained in O}, where N(x,e) is the interval of radius e centered at x. (2) To see that Rx>0 for all x, suppose there is an x where Rx=0. Then N(x,Rx)={x}, and that shows that x is not an interior point of O, contradicting that O is open, for else every point of O would be interior. (3) For x in O, let Ox=union ( N(y,Ry) : x in N(y,Ry) ). Since the union doesn't depend on how we order the y's for the index set, Ox=Oy for any y in Ox. (4) O=union ( Ox : x in O ), since each x appears at least in Ox. (5) Each Ox is an interval: to see this, notice that Ox is connected, for each x. This is true, since for any y in Ox, x is in N(y,Ry) and N(y,Ry) is a subset of Ox; so if Ox were disconnected then one of these intervals N(y,Ry) would be, which contradicts that an interval is connected. (6) Now that we see that Ox is connected, note that the closure of Ox is compact, since Ox is bounded, as a subspace of a bounded space. So, using the extreme value theorem, let a=min { closure Ox } and b=max { closure Ox } (these are the min and max of the identity function on closure (Ox)). To see that Ox=(a,b), note that Ox contained in (a,b) since for all y in Ox, a

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