Question

The distance from a point \(\mathbf{X}_{0}\) to a nonempty set \(S\) is defined by $$ \operatorname{dist}\left(\mathbf{X}_{0}, S\right)=\inf \left\\{\left|\mathbf{X}-\mathbf{X}_{0}\right| \mid \mathbf{X} \in S\right\\} $$ (a) Prove: If \(S\) is closed and \(\mathbf{X}_{0} \in \mathbb{R}^{n},\) there is a point \(\overline{\mathbf{X}}\) in \(S\) such that $$ \left|\overline{\mathbf{X}}-\mathbf{X}_{0}\right|=\operatorname{dist}\left(\mathbf{X}_{0}, S\right) $$ $$ C_{m}=\left\\{\mathbf{X} \mid \mathbf{X} \in S \text { and }\left|\mathbf{X}-\mathbf{X}_{0}\right| \leq \operatorname{dist}\left(\mathbf{X}_{0}, S\right)+1 / m\right\\}, \quad m \geq 1 $$ (b) Show that if \(S\) is closed and \(\mathbf{X}_{0} \notin S,\) then \(\operatorname{dist}\left(\mathbf{X}_{0}, S\right)>0\) (c) Show that the conclusions of (a) and (b) may fail to hold if \(S\) is not closed.

Short answer

Question: Prove that for a closed, non-empty set \(S\) and a point \(\mathbf{X}_0\), there exists a point \(\overline{\mathbf{X}}\) in \(S\) such that the distance between \(\overline{\mathbf{X}}\) and \(\mathbf{X}_{0}\) is equal to \(\operatorname{dist}(\mathbf{X}_{0}, S)\). Also, if \(\mathbf{X}_0 \notin S\), then \(\operatorname{dist}(\mathbf{X}_0,S)>0\). Provide counterexamples for a non-closed set. Answer: For a closed set \(S\), there exists a point \(\overline{\mathbf{X}}\) in \(S\) such that the distance between \(\overline{\mathbf{X}}\) and \(\mathbf{X}_{0}\) is equal to \(\operatorname{dist}(\mathbf{X}_{0}, S)\). If \(\mathbf{X}_0 \notin S\), then \(\operatorname{dist}(\mathbf{X}_0,S)>0\). Counterexamples for a non-closed set are: (i) Set \(S=(0,1)\) and \(\mathbf{X}_0=0\), where no point in \(S\) has the minimum distance from \(\mathbf{X}_0\). (ii) Set \(S=(0,1)\) and \(\mathbf{X}_0=1\), where \(\operatorname{dist}(\mathbf{X}_0,S)=0\), even though \(\mathbf{X}_0 \notin S\).

Step-by-step solution

Part (a): Proof of existence of \(\overline{\mathbf{X}}\) in closed set S

Consider the definition of \(C_{m}=\left\\{\mathbf{X} \mid \mathbf{X} \in S \text { and }\left|\mathbf{X}-\mathbf{X}_{0}\right| \leq \operatorname{dist}\left(\mathbf{X}_{0}, S\right)+\frac{1}{m}\right\\}, \quad m \geq 1\). Since \(S\) is closed and non-empty, we know that \(C_{m}\) is also non-empty for all \(m \geq 1\). For all \(m \geq 1\), let \(\mathbf{X}_{m} \in C_{m}\). By the definition of \(C_{m}\), we have: $$ \left|\mathbf{X}_{m}-\mathbf{X}_{0}\right| \leq \operatorname{dist}\left(\mathbf{X}_{0}, S\right) + \frac{1}{m} $$ Taking the limit as \(m \to \infty\), we get: $$ \lim_{m\to\infty}\left|\mathbf{X}_{m}-\mathbf{X}_{0}\right| \leq \operatorname{dist}\left(\mathbf{X}_{0}, S\right) $$ On the other hand, since \(\mathbf{X}_{m} \in S\), we also have: $$ \operatorname{dist}\left(\mathbf{X}_{0}, S\right) \leq \lim_{m\to\infty}\left|\mathbf{X}_{m}-\mathbf{X}_{0}\right| $$ Hence, we get: $$ \lim_{m\to\infty}\left|\mathbf{X}_{m}-\mathbf{X}_{0}\right| = \operatorname{dist}\left(\mathbf{X}_{0}, S\right) $$ Since \(S\) is closed, we know that the limit of a convergent sequence in \(S\) must also exist in \(S\). Therefore, there exists \(\overline{\mathbf{X}} \in S\) such that \(\lim_{m\to\infty}\mathbf{X}_{m} = \overline{\mathbf{X}}\). Hence, we can say: $$ \left|\overline{\mathbf{X}}-\mathbf{X}_{0}\right|=\operatorname{dist}\left(\mathbf{X}_{0}, S\right) $$

Part (b): Distance is positive if \(\mathbf{X}_{0} \notin S\) and \(S\) is closed

Assume that \(\operatorname{dist}\left(\mathbf{X}_{0}, S\right) = 0\). Then, according to the definition of the distance, there exists a sequence of points in \(S\), say \(\{\mathbf{Y}_{n}\}_{n=1}^{\infty}\), such that the following holds: $$ \lim_{n\to\infty}\left|\mathbf{Y}_n - \mathbf{X}_{0}\right| = 0 $$ This means, \(\lim_{n\to\infty}\mathbf{Y}_n = \mathbf{X}_{0}\). Since \(S\) is closed, it contains all its limit points. Thus, \(\mathbf{X}_{0}\) must be in \(S\), which contradicts the initial condition that \(\mathbf{X}_{0} \notin S\). Therefore, our assumption that \(\operatorname{dist}\left(\mathbf{X}_{0}, S\right) = 0\) must be false. Hence, we can conclude that \(\operatorname{dist}\left(\mathbf{X}_{0}, S\right) > 0\).

Part (c): Counterexamples for non-closed set \(S\)

(i) Consider a set \(S = (0,1) \subset \mathbb{R}\), which is an open interval and, therefore, not closed. Let \(\mathbf{X}_{0} = 0\). Since \(\mathbf{X}_{0}\) is not contained in \(S\), the distance from \(\mathbf{X}_{0}\) to \(S\) is non-zero. However, for any \(\epsilon > 0\), there is a point in \(S\) that is less than \(\epsilon\) away from \(\mathbf{X}_{0}\). But there is no point in \(S\) with the minimum distance from \(\mathbf{X}_{0}\), as we approach \(\mathbf{X}_{0}\) without reaching it. Thus, the conclusion of part (a) fails to hold in this case. (ii) Now consider the same set \(S = (0,1)\) and \(\mathbf{X}_{0} = 1\). In this case, we have \(\mathbf{X}_{0} \notin S\), but \(\operatorname{dist}(\mathbf{X}_{0},S) = 0\), since there is a sequence of points in \(S\) converging to \(\mathbf{X}_{0}\). Thus, the conclusion of part (b) fails to hold in this case.

Key concepts

Closed Set

In real analysis, a closed set is fundamental to understanding concepts like convergence and limits within a mathematical space. A set, typically within a real or complex number space, is considered closed if it contains all of its limit points.

Imagine standing on the edge of a pool—the water representing the closed set. Being closed means that if you dip your toes into the water, those points on your toes are included within the set—you're still in contact with the water. In contrast, if the set were open, dipping your toes might not always mean they're part of the set, like dipping your toes over the pool edge without touching the water.

Here's why this matters: closed sets have a cozy property where any sequence of points within the set that gets closer and closer to some point will actually reach a point within the set itself. This contrasts with an open set, where sequences can approach a point that isn't included in the set, much like approaching the water's surface without breaking through.

Point-to-Set Distance

The point-to-set distance is a measure of the proximity of a solitary point to an entire set of points. It's a bit like measuring the shortest distance from a lighthouse to the shoreline—what's the closest point of land to the lighthouse? Mathematically, this distance is defined as the infimum (or the 'greatest lower bound') of the distances between the point and each point in the set.

If you're picturing that lighthouse, it might stand alone on a rocky outcrop. The infimum is the shortest distance between the lighthouse and every possible point along the coast. In defining this distance, we can understand whether the lighthouse is part of the landmass (like if it's built on the shore) or stands separate from it. This shortest distance plays a significant role in optimization and analysis, as finding the minimum or closest point can be a crucial step in solving broader mathematical problems.

Convergent Sequence

A convergent sequence is a lineup of points that get increasingly close to a specific point, known as the sequence's limit. Picture a flock of birds settling down: no matter how much they circle, eventually, they all land on one tree. That tree is analogous to the limit of the convergence.

In technical terms, for any amount of 'closeness' (epsilon), you can find a spot in the sequence past which all points are at least that close to the limit point. Now, this gets interesting in closed sets because if the sequence lives inside a closed set, the point they're getting close to also has to be within the same set—none of those birds can land on a different tree. Convergence is a cornerstone of mathematical analysis since it helps characterize functions, series, and integrals and forms the bedrock of understanding continuity and limits.