Is the Union of a Bounded Set and Its Accumulation Points Also Bounded?

[Argonlight]

The homework question is this:
Prove If D is a bounded subset of R then D bar = D U D’ is also bounded where D’ is the set of accumulation points of D.

What is a general outline of a proof?

 

[joeboo]

It suffices to show that D' is bounded, as the union of 2 bounded sets is bounded.

If D is bounded, then it is contained in some finite interval, ( - N, N ). If D' is not bounded, then we can find an element, x, of D' outside of ( -N, N ), and taking a suitable neighborhood around x ( of radius less than |x| - N ), we see that it is disjoint from D ( as it is disjoint from ( -N, N ) ). Therefore, x is not an accumulation point of D. Contradiction

 

[M98Ranger]

A general outline of a proof for this statement would be as follows:

1. Start by assuming that D is a bounded subset of R. This means that there exists some M > 0 such that for all x in D, |x| < M.

2. Next, we need to show that D U D' is bounded. To do this, we need to find a value N > 0 such that for all y in D U D', |y| < N.

3. Since D' is the set of accumulation points of D, every point in D' is either in D or is a limit point of D. This means that for any point z in D', there exists a sequence (xn) in D such that xn -> z as n -> infinity.

4. By the definition of convergence, we know that for any epsilon > 0, there exists some N such that for all n > N, |xn - z| < epsilon. This means that for all n > N, z - epsilon < xn < z + epsilon.

5. Since D is bounded, we know that |xn| < M for all n. Combining this with the previous inequality, we get z - epsilon < xn < z + epsilon < M. This holds for all n > N, so it also holds for the limit point z.

6. Therefore, we can say that for any limit point z in D', there exists some M' = max{|z| + epsilon, M} such that |z| < M'. This holds for all limit points in D', so we can say that D' is also bounded.

7. Finally, since D U D' is the union of two bounded sets, it is also bounded. We can choose N = max{M, M'} as our bound for D U D'.

8. Thus, we have shown that if D is a bounded subset of R, then D U D' is also bounded.