1 Answer | Add Yours
Let `S=QQ.` The closure of `QQ` is the union of `QQ` and its limit points. But every real number is a limit point of `QQ,` since `QQ` is dense in `RR` , so ``
cl `QQ` =`RR` .
The boundary of `RR` is the set of all points `x` in `RR` such that every neighborhood of `x` contains a point in `RR` and a point not in `RR`. But no neighborhood of a point on the number line contains a point not in `RR` , so the boundary of `RR` is the empty set. So
bd (cl `QQ` )=`O/` .
Now, the boundary of `QQ` is `RR` , because for any `x inRR` , every neighborhood of `x` contains both rational (elements of `QQ` ) and irrational (not elements of `QQ` ) numbers. Again, this is due to the fact that `QQ` is dense in `RR` (see Theorem 8 in the link for a proof). Thus we have the counterexample
bd(cl `QQ` )`!=` ` ` bd `QQ.`
We’ve answered 317,691 questions. We can answer yours, too.Ask a question