Understanding Metric Spaces, Closed Sets, and Complete Sets

The question asks about a property of a closed subset within a metric space. It states "Let (X, d) be a metric sparse space and let B be a subset of X then if B is closed then B is also ______." The term "metric sparse space" is not standard terminology in topology or analysis. It is highly probable that "sparse" is a typographical error and the question intended to refer to a "complete metric space". We will proceed by analyzing the question under the assumption that X is a complete metric space, as the provided answer aligns with a well-known theorem in this context.

What is a Metric Space?

A metric space is a set X equipped with a function d (called a metric or distance function) that satisfies the following properties for all x, y, z in X:

1. Non-negativity: $d(x, y) \ge 0$
2. Identity of indiscernibles: $d(x, y) = 0$ if and only if $x = y$
3. Symmetry: $d(x, y) = d(y, x)$
4. Triangle inequality: $d(x, z) \le d(x, y) + d(y, z)$

What is a Closed Subset?

A subset B of a metric space (X, d) is called closed if it contains all its limit points. Equivalently, B is closed if its complement X \ B is an open set in X.

What is a Complete Metric Space?

A metric space (X, d) is called complete if every Cauchy sequence in X converges to a point in X.

• A sequence $(x_n)$ in X is a Cauchy sequence if for every $\epsilon > 0$, there exists an integer N such that for all m, n > N, $d(x_m, x_n) < \epsilon$.
• A sequence $(x_n)$ in X converges to a point x $\in$ X if for every $\epsilon > 0$, there exists an integer N such that for all n > N, $d(x_n, x) < \epsilon$.

Analyzing the Relationship Between Closed Sets and Complete Sets

The crucial theorem related to this question (assuming X is complete) is:

Theorem: Every closed subset of a complete metric space is complete.

Let's understand why this theorem is true. Suppose (X, d) is a complete metric space, and B is a closed subset of X. Let $(b_n)$ be a Cauchy sequence in B. Since B is a subset of X, $(b_n)$ is also a Cauchy sequence in X. Because X is complete, the sequence $(b_n)$ converges to some point, say $x$, in X. Since B is closed and $(b_n)$ is a sequence in B converging to $x$, the limit point $x$ must belong to B. Therefore, every Cauchy sequence in B converges to a point in B, which means B is complete with the induced metric from X.

Evaluating the Options

Given the likely interpretation that (X, d) is a complete metric space and B is a closed subset:

1. Unbounded: A closed set in a complete metric space is not necessarily unbounded. For example, the closed interval [0, 1] is closed and bounded in the complete metric space ℝ.
2. Complete: As explained by the theorem, a closed subset of a complete metric space is complete. This aligns with the provided correct answer.
3. Dense: A closed set is not necessarily dense. A set B is dense in X if its closure $\bar{B}$ is equal to X. If B is closed, $\bar{B} = B$. So B is dense if and only if B = X. A proper closed subset (like {0} in ℝ) is closed but not dense.
4. Bounded: A closed set in a complete metric space is not necessarily bounded. For example, the set of integers $\mathbb{Z}$ is closed in the complete metric space ℝ but it is not bounded.

Based on the analysis and the standard theorem, if we assume (X, d) is a complete metric space, then a closed subset B of X is also complete.

Thus, if B is closed and (X, d) is a complete metric space, then B is also complete.

Revision Table: Metric Space Concepts

Additional Information on Completeness

Completeness is a very important property in metric spaces. It guarantees that there are "no holes" in the space in terms of convergence of sequences. Here are some related facts:

• The space of real numbers ℝ with the standard metric $d(x,y) = |x-y|$ is complete.
• The space of rational numbers ℚ with the standard metric is not complete. For example, a sequence of rational numbers converging to $\sqrt{2}$ (which is irrational) is a Cauchy sequence in ℚ but does not converge to a point in ℚ.
• A subset B of a metric space (X, d) is complete if and only if it is closed in X and the metric space (B, d$_B$) is complete, where d$_B$ is the restricted metric.
• The theorem used in the solution, "A closed subset of a complete metric space is complete," is a fundamental result. However, the converse is not true: a complete subset of a metric space is always closed, regardless of whether the parent space X is complete.
• Compact metric spaces are always complete.

Assuming "sparse" was a typo for "complete", the property of B being closed implies it inherits the completeness from the parent space X.