Question

If \(A \subset R^{k}\) and \(B \subset R^{k}\), define \(A+B\) to be the set of all sums \(\mathbf{x}+\mathbf{y}\) with \(\mathbf{x} \in A\) \(\mathbf{y} \in \boldsymbol{B}\) (a) If \(K\) is compact and \(C\) is closed in \(R^{k}\), prove that \(K+C\) is closed. Hint: Take \(z \notin K+C\), put \(F=z-C\), the set of all \(\mathbf{z}-\mathbf{y}\) with \(\mathbf{y} \in \boldsymbol{C}\). Then \(K\) and \(F\) are disjoint. Choose \(\delta\) as in Exercise 21. Show that the open ball with center \(\mathrm{z}\) and radius \(\delta\) does not intersect \(K+C\). (b) Let \(\alpha\) be an irrational real number. Let \(C_{1}\) be the set of all integers, let \(C_{2}\) be the set of all \(n \alpha\) with \(n \in C_{1}\). Show that \(C_{1}\) and \(C_{2}\) are closed subsets of \(R^{1}\) whose sum \(C_{1}+C_{2}\) is not closed, by showing that \(C_{1}+C_{2}\) is a countable dense subset of \(R^{1} .\)

Short answer

Question: Prove that if K is a compact set and C is a closed set in R^k, then K+C is a closed set. Also, show that C_1+C_2 is not closed, where C_1 and C_2 are closed subsets of R^1 defined by C_1 being the set of integers and C_2 being the set of integer multiples of an irrational real number alpha. Answer: To prove that K+C is closed, we showed that for any z not in K+C, there exists an open ball centered at z with radius delta that does not intersect K+C. For the second part, we showed that C_1+C_2 is a countable dense subset of R^1 by proving that every open interval in R^1 contains at least one element from C_1+C_2. Since a countable dense subset cannot be closed, C_1+C_2 is not closed.

Step-by-step solution

Define the given sets and their sum

Given a compact set \(K \subset R^k\) and a closed set \(C \subset R^k\), the sum of the sets \(K+C\) is defined as the set of all sums \(\mathbf{x}+\mathbf{y}\) with \(\mathbf{x} \in A\) and \(\mathbf{y} \in B\).

Define the set F

For \(z \notin K+C\), define \(F = z-C\) as the set of all \(\mathbf{z}-\mathbf{y}\) with \(\mathbf{y} \in C\). Thus, the sets \(K\) and \(F\) are disjoint.

Choose suitable delta

As \(K\) is compact, for every open cover of \(K\), there exists a finite subcover. This implies that there exists a minimum distance \(\delta > 0\) between the sets \(K\) and \(F\).

Prove the open ball does not intersect K+C

Consider the open ball \(B(z, \delta)\) centered at \(z\) with radius \(\delta\). Since \(\delta > 0\) is the minimum distance between \(K\) and \(F\), \(B(z, \delta)\) does not intersect the set \(K+C\).

Conclude that K+C is closed

Since \(B(z, \delta)\) does not intersect \(K+C\), and this holds true for all \(z \notin K+C\), we can conclude that \(K+C\) is a closed set in \(R^k\). (b) Prove that \(C_1+C_2\) is a countable dense subset of \(R^1\).

Define the closed subsets C1 and C2

Let \(\alpha\) be an irrational real number. \(C_1\) is the set of all integers, and \(C_2\) is the set of all \(n \alpha\) with \(n \in C_1\).

Show that the sum C1+C2 is countable

We have to show that the sum \(C_1+C_2\) is countable. Since both \(C_1\) and \(C_2\) are countable, their sum \(C_1+C_2\) is also countable.

Show that C1+C2 is dense in R1

To show that \(C_1+C_2\) is dense in \(R^1\), we have to show that every open interval in \(R^1\) contains at least one element of \(C_1+C_2\). We can do this by proving that for every real number \(x\), there exists integers \(m\) and \(n\) such that \(x-\epsilon < m+n\alpha < x+\epsilon\) for any \(\epsilon>0\). Since \(\alpha\) is irrational, the set \(\{m\alpha \mod 1 : m\in \mathbb{Z}\}\) is dense in \((0,1)\), which implies that the sum \(C_1+C_2\) is dense in \(R^1\).

Conclude that the sum C1+C2 is not closed

Since \(C_1+C_2\) is a countable dense subset of \(R^1\), it cannot be a closed set, as the closure of a countable dense subset in \(R^1\) is uncountable (in this case, it is \(R^1\) itself). Therefore, we have shown that the sum \(C_1+C_2\) is not closed.

Key concepts

Compact Sets

In mathematical analysis, a compact set possesses an intuitive property: it behaves, in a sense, like a finite set. A set is compact if it is both closed and bounded. This means that:

• Closed: Every limit point of the set is contained within the set.
• Bounded: There exists some real number that is larger than the magnitude of any point in the set.

In the context of real numbers, this implies that you can "cover" the set with a finite number of open intervals and not miss any points of the set. Compactness is crucial because it allows for powerful results like the Heine-Borel theorem, which states that in Euclidean space, a set is compact if and only if it's closed and bounded.
Compact sets are important in the given exercise as their properties help derive results about the closure of set sums.

Closed Sets

Closed sets in mathematical analysis have a straightforward characteristic: they include all their limit points. Essentially, if you could get close "up to the edge" of a set, that edge must be part of the closed set.
To picture this, imagine a set defined by a circle in two-dimensional space. If the boundary of the circle, as well as the points inside, belong to the set, the set is closed. Key properties of closed sets include:

• The intersection of closed sets is closed.
• The complement of an open set is closed.
• The closure of any set (i.e., adding all its limit points) results in a closed set.

In the given exercise, knowing that a set is closed helps determine properties like closure and disjointness when dealing with other sets.

Dense Sets

The concept of dense sets touches upon the idea of "filling in" a space. A subset of a metric space is considered dense if every point in the space can be approached arbitrarily closely by points from the subset.
If you think about rational numbers within the real number line, they form a dense set because between any two real numbers, you can find a rational number. For a set to be dense means that:

• In every open interval, no matter how small, there is at least one point from the set.
• Dense sets are critical when discussing the closure of these sets.

In the second part of the exercise, proving that sums of certain sets are dense shows why their closure would eventually encompass the entirety of the space such as \( R^1 \).

Countable Sets

Countable sets in the realm of mathematical analysis refer to sets whose elements can be matched one-to-one with the natural numbers. This means that a countable set has the same size (cardinality) as some subset of the natural numbers. These can be either finite sets or countably infinite like the set of all integers or rational numbers. The key features include:

• If you can list the elements of a set one by one and never miss any of them, the set is countable.
• The union of two countable sets is countable.
• Countable sets are vital because they help classify the "size" of infinity of various sets.

In the provided exercise, understanding that the sum of two countable, dense sets remains countable illustrates why the new set might not be closed in a larger metric space.