Question

The distance between two nonempty sets \(S\) and \(T\) is defined by $$ \operatorname{dist}(S, T)=\inf \\{|\mathbf{X}-\mathbf{Y}| \mid \mathbf{X} \in S, \mathbf{Y} \in T\\} $$ (a) Prove: If \(S\) is closed and \(T\) is compact, there are points \(\overline{\mathbf{X}}\) in \(S\) and \(\overline{\mathbf{Y}}\) in \(T\) such that $$ |\overline{\mathbf{X}}-\overline{\mathbf{Y}}|=\operatorname{dist}(S, T) $$ (b) Under the assumptions of (a), show that \(\operatorname{dist}(S, T)>0\) if \(S \cap T=\emptyset\). (c) Show that the conclusions of (a) and (b) may fail to hold if \(S\) or \(T\) is not closed or \(T\) is unbounded.

Short answer

#Short Answer# We can find points X_bar ∈ S and Y_bar ∈ T with distance equal to the distance between the sets S and T if set S is closed and set T is compact by using the properties of closed and compact sets, as well as the definition of distance between sets. When S and T are disjoint, their distance is positive. However, if either set is not closed or if set T is unbounded, the conclusions of the previous parts may fail. Counterexamples for non-closed sets and unbounded sets can be provided to illustrate these cases.

Step-by-step solution

Use the definition of distance between sets

Given the distance between sets S and T: $$ \operatorname{dist}(S, T) = \inf \\{|\mathbf{X}-\mathbf{Y}| \mid \mathbf{X} \in S, \mathbf{Y} \in T\\} $$ We're required to prove the existence of points X_bar ∈ S and Y_bar ∈ T with distance equal to dist(S, T). We will use the properties of closed and compact sets to find these points.

Finding a sequence of points and using properties of compact sets

Let us choose a sequence of points {\(X_n\)} ⊆ S and {\(Y_n\)} ⊆ T such that $$ \lim_{n \to \infty} |X_n - Y_n| = \operatorname{dist}(S, T) $$ Since T is compact, there exists a convergent subsequence {\(Y_{n_k}\)} with limit point Y_bar ∈ T. Now, consider the sequence {\(X_{n_k}\)}.

Use the properties of closed sets

As S is closed, it contains all its limit points. Therefore, {\(X_{n_k}\)} has a convergent subsequence {\(X_{n_{k_j}}\)} with limit point X_bar ∈ S.

Prove the existence of X_bar and Y_bar

Using the above sequences and subsequences, we have: $$ \lim_{j \to \infty} |X_{n_{k_j}} - Y_{n_{k_j}}| = \operatorname{dist}(S, T) $$ Taking the limit as j approaches infinity, we obtain: $$ |X_\text{bar} - Y_\text{bar}| = \operatorname{dist}(S, T) $$ Thus, we have proved the existence of points X_bar ∈ S and Y_bar ∈ T. (b) Proving dist(S, T) > 0 if S ∩ T = ∅

Use the result from part (a)

From part (a), we know that there exist X_bar ∈ S and Y_bar ∈ T such that: $$ |X_\text{bar} - Y_\text{bar}| = \operatorname{dist}(S, T) $$

Prove that dist(S, T) is positive

Since S ∩ T = ∅ (i.e., S and T are disjoint), we know that X_bar ≠ Y_bar. Therefore, the distance between X_bar and Y_bar is positive: $$ |X_\text{bar} - Y_\text{bar}| > 0 $$ Thus, we can conclude that if S ∩ T = ∅: $$ \operatorname{dist}(S, T) > 0 $$ (c) Counterexamples for failing conclusions of (a) and (b)

Counterexample for non-closed set

Let S = (0, 1) and T = {1}. Here, S is not closed and T is bounded. Although the distance between the sets is dist(S, T) = 0, S ∩ T = ∅. This counters the conclusion of (b).

Counterexample for unbounded set

Let S = \(\mathbb{Z}\) and T = \(\{x \in \mathbb{R} \mid x^2 > 2\}\). Here, S is closed but T is unbounded. The distance between the sets is dist(S, T) = 0, but S ∩ T = ∅. This also counters the conclusion of (b).

Key concepts

Compact Sets

In the context of real analysis, compact sets are crucial because they blend the properties of being closed and bounded. Think of them as the 'best-behaved' sets in terms of convergence and continuity. For any sequence in a compact set, you can always find a subsequence that converges to a point within the same set. This characteristic is essential in the proof of the distance between sets, specifically in ensuring that the sequence { \(\{Y_n\}\) } converges to a point in T.

For a real understanding, picture a compact set as a secure bubble—no matter how you bounce around inside the bubble (how the sequence behaves), you will eventually settle down to a point within the bubble (a convergent subsequence). This consistency provides a strong foundation when exploring the relationships between sets, such as in our exercise.

Closed Sets

While closed sets may seem less stringent than compact sets, they play a pivotal role in defining the distance between two sets. A closed set is one that contains all its limit points, meaning that if you were to step infinitely close to the edge of a closed set, you would never fall out—they hold on to their boundaries tightly.

Using closed sets is analogous to drawing with a pen that has an ink-tight cap; once you stop drawing, no extra ink escapes—the line stops exactly where you intended. This aids in the proof of our exercise by guaranteeing that the limit points identified within the sequence from set S remain within S, thus properly defining the distance to T.

Infimum of a Set

The infimum of a set, often abbreviated as 'inf,' is the greatest lower bound of the set. It is the smallest value that is not less than any other value in the set. The concept of infimum becomes the deciding factor in determining the distance between sets S and T because it defines the least possible separation between any two points from each set.

To visualize the infimum, think of the floor in a room full of bouncy balls—no matter how small the bounce, the balls cannot go below the floor. The infimum is crucial in our exercise, forming the baseline measurement from which we define the actual distance between S and T.

Limit Points

A limit point of a set is a point that can be approached by a sequence from within the set as closely as desired, without necessarily being a point of the sequence itself. It's like being at a busy street corner; even if you're not part of the passing crowd, individuals from the crowd can continuously approach you.

In our exercise, identifying limit points of certain sequences within the sets S and T allows us to conclude that there are points as close as the infimum distance apart. This recognition is instrumental in proving that the distance between the two sets can be realized by actual points, bringing the abstract definition of 'inf' into the concrete realm of distances between elements.