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

• Argonlight
In summary, to prove that the set D bar = D U D' is also bounded if D is a bounded subset of R, it suffices to show that D' is bounded. This can be done by considering an element x in D' that is outside of the finite interval (-N, N) where D is contained, and showing that it is disjoint from D. This leads to a contradiction, proving that D' must be 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?

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

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.

## What does "Prove D U D' is bounded" mean?

The statement "Prove D U D' is bounded" is asking for a proof that the set D U D' is bounded. This means that there exists a number M such that every element in D U D' is less than or equal to M.

## How can I prove that D U D' is bounded?

To prove that D U D' is bounded, you can use the definition of a bounded set. This means showing that there exists a number M such that all elements in D U D' are less than or equal to M.

## What does it mean for a set to be bounded?

A set is bounded if there exists a number M such that all elements in the set are less than or equal to M. This means that the values in the set do not go to infinity and are limited by a certain value.

## Why is it important to prove that D U D' is bounded?

Proving that D U D' is bounded is important because it shows that the set has a finite limit and does not go to infinity. This can be useful in many mathematical and scientific calculations and proofs.

## Can you give an example of a bounded set?

One example of a bounded set is the set of all real numbers between 0 and 1. In this set, the number 1 is the upper bound, as all values in the set are less than or equal to 1.

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