Show that any open interval (a, b) in R is an open ball. Is R an open ball in R

To show that any open interval (a, b) in R is an open ball, we need to prove that for any point x in (a, b), there exists a positive radius r such that the open ball B(x, r) is completely contained within (a, b).

Let's consider an arbitrary point x in (a, b). Since x is in the open interval (a, b), we know that a < x < b.

Now, let's define the radius r as the minimum of the distances between x and the endpoints a and b, i.e., r = min(x - a, b - x). Since a < x < b, both x - a and b - x are positive, so r is positive.

Next, we need to show that the open ball B(x, r) is completely contained within (a, b).

Let y be any point in B(x, r). By definition, this means that the distance between y and x is less than r. Therefore, |y - x| < r.

Since r = min(x - a, b - x), we have two cases to consider:

1. If y < x, then |y - x| = x - y. Since |y - x| < r = x - a, we have x - y < x - a, which implies y > a. But since y < x, we have a < y < x, which means y is in the open interval (a, x).

2. If y > x, then |y - x| = y - x. Since |y - x| < r = b - x, we have y - x < b - x, which implies y < b. But since y > x, we have x < y < b, which means y is in the open interval (x, b).

In both cases, we have shown that y is in the open interval (a, b). Therefore, any point y in B(x, r) is also in (a, b), which means B(x, r) is completely contained within (a, b).

Hence, we have shown that any open interval (a, b) in R is an open ball.

Now, let's consider R itself. R is not an open ball in R because an open ball requires a center point and a positive radius. However, R is not bounded, so we cannot find a specific center point and radius that would define an open ball contained within R.