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Chapter 6: Problem 12

Suppose \(V\) is a normed vector space. Prove that the closure of each subspace of \(V\) is a subspace of \(V\).

Short Answer

Expert verified
To prove that the closure \( \overline{U} \) of a subspace \( U \) of a normed vector space \( V \) is also a subspace of \( V \), we need to show that \( \overline{U} \) is closed under addition and scalar multiplication. We showed that, for any \( x, y \in \overline{U} \) and \( c \in \mathbb{F} \), \( x+y \in \overline{U} \) and \( cx \in \overline{U} \) by proving that the sequences \( (x_n + y_n)_{n\in\mathbb{N}} \) and \( (cx_n)_{n\in\mathbb{N}} \), converging respectively to \( x+y \) and \( cx \), have their elements in \( U \). Thus, the closure of each subspace of a normed vector space is a subspace of that vector space.

Step by step solution

01

Define Closure of a Subspace

Let \(U\) be a subspace of normed vector space \(V\). We want to prove that the closure of \(U\), denoted as \(\overline{U}\), is also a subspace of \(V\). By definition, the closure \(\overline{U}\) is the smallest closed subset of \(V\) containing \(U\).
02

Property of Addition

We need to show that the closure \(\overline{U}\) is closed under vector addition. Let \(x, y \in \overline{U}\), meaning \(x\) and \(y\) are limit points of sequences \((x_n)_{n\in\mathbb{N}}\) and \((y_n)_{n\in\mathbb{N}}\) respectively, with \(x_n, y_n \in U\). We want to prove that \(x+y \in \overline{U}\). Consider the sequence \((x_n + y_n)_{n\in\mathbb{N}}\). Since \(U\) is a subspace, for all \(n \in \mathbb{N}\), \(x_n+y_n \in U\). Now, as \(x_n \to x\) and \(y_n \to y\), by the properties of limits, we know that \(x_n+y_n \to x+y\). Therefore, \(x+y\) is a limit point of a sequence in \(U\), meaning \(x+y \in \overline{U}\).
03

Property of Scalar Multiplication

Next, we need to show that the closure \(\overline{U}\) is closed under scalar multiplication. Let \(x \in \overline{U}\) and \(c \in \mathbb{F}\) (where \(\mathbb{F}\) is the scalar field). Our goal is to prove that \(cx \in \overline{U}\). As \(x\) is in the closure, it is the limit point of a sequence \((x_n)_{n\in\mathbb{N}}\) with \(x_n \in U\). Consider the sequence \((cx_n)_{n\in\mathbb{N}}\). Since \(U\) is a subspace, we know that for all \(n \in \mathbb{N}\), \(cx_n \in U\). As \(x_n \to x\), by the properties of limits, we have that \(cx_n \to cx\). Therefore, \(cx\) is a limit point of a sequence in \(U\), meaning \(cx \in \overline{U}\).
04

Conclusion

We have shown that the closure of a subspace \(U\) of normed vector space \(V\), denoted as \(\overline{U}\), is closed under addition and scalar multiplication. Therefore, the closure of each subspace of a normed vector space is a subspace of that vector space. This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Closure of a Subspace
In the world of mathematics, particularly in normed vector spaces, the concept of closure is crucial. When we talk about the "closure of a subspace," we're referring to the smallest closed subset that contains the entire subspace. Imagine the closure as a snug boundary around a subspace that stretches just enough to cover all the points in the subspace, including its limit points.

The closure of a subspace ensures that it isn't just about being inside the space itself, but also about encompassing all the limit points that could approach as closely as desired. This concept is immensely important when dealing with sequences and functional analysis. Here, it's not just about where points are, but also about how sequences of points can converge towards other points within the space.
Vector Addition
When discussing vector spaces, one must understand vector addition deeply. This operation is not only simple but also foundational. In a subspace, every pair of vectors should combine under addition to form another vector that is still within the same subspace.

By ensuring vector addition works within the closure, you validate that, even as vectors reach towards the boundaries of the subspace, their sums remain within the closure. For instance, if two sequences in a subspace converge to vectors, then the sequence of their sums should also converge to a vector within the closure. This characteristic ensures that closure under vector addition holds strong, meaning that the compound of any two vectors stays within the set.
Scalar Multiplication
Scalar multiplication in vector spaces involves taking vectors and stretching or shrinking them by a scalar from the underlying field. In a subspace and its closure, scalar multiplication must not push the vector out of bounds.

If vector \( x \) belongs to the closure of a subspace, multiplying \( x \) by any scalar from the field should keep the result within the closure. Essentially, if a sequence converges to a vector in the closure, then the sequence of that vector multiplied by any scalar should converge to a point within the closure too. This feature guarantees that scalar multiplication doesn鈥檛 disrupt the subspace properties of the closure.
Subspace Properties
Subspace properties are foundational rules that every subspace must follow to maintain its structure within a larger vector space. These properties are essential for analyzing vector spaces and distinguishing valid subspaces.

Every subspace must:
  • Include the zero vector, or origin, ensuring it's part of the space.
  • Be closed under vector addition, meaning the sum of any two vectors stays within the subspace.
  • Be closed under scalar multiplication, ensuring any vector multiplied by a scalar stays within the subspace.
These properties are crucial as they not only define the subspace but also allow for the logical extension of results and operations within the vector space. They act as a framework for understanding how subspaces and their closures fit perfectly in the grand landscape of vector spaces.

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Most popular questions from this chapter

Suppose \(V\) is a separable normed vector space. Explain how the Hahn-Banach Theorem (6.69) for \(V\) can be proved without using any results (such as Zorn's Lemma) that depend upon the Axiom of Choice.

Suppose \((X, \mathcal{S})\) is a measurable space and \(f, g: X \rightarrow \mathbf{C}\) are \(\mathcal{S}\) -measurable. Prove that (a) \(f+g, f-g,\) and \(f g\) are \(\mathcal{S}\) -measurable functions; (b) if \(g(x) \neq 0\) for all \(x \in X,\) then \(\frac{f}{g}\) is an \(\mathcal{S}\) -measurable function.

Suppose \(\left(U, d_{U}\right),\left(V, d_{V}\right),\) and \(\left(W, d_{W}\right)\) are metric spaces. Suppose also that \(T: U \rightarrow V\) and \(S: V \rightarrow W\) are continuous functions. (a) Using the definition of continuity, show that \(S \circ T: U \rightarrow W\) is continuous. (b) Using the equivalence of \(6.11(\) a) and \(6.11(\mathrm{~b})\), show that \(S \circ T: U \rightarrow W\) is continuous. (c) Using the equivalence of \(6.11(\) a) and \(6.11(\mathrm{c}),\) show that \(S \circ T: U \rightarrow W\) is continuous.

The double dual space of a normed vector space is defined to be the dual space of the dual space. If \(V\) is a normed vector space, then the double dual space of \(V\) is denoted by \(V^{\prime \prime} ;\) thus \(V^{\prime \prime}=\left(V^{\prime}\right)^{\prime} .\) The norm on \(V^{\prime \prime}\) is defined to be the norm it receives as the dual space of \(V^{\prime}\). Suppose \(V\) is an infinite-dimensional normed vector space. Show that there is a convex subset \(U\) of \(V\) such that \(\bar{U}=V\) and such that the complement \(V \backslash U\) is also a convex subset of \(V\) with \(\overline{V \backslash U}=V\). [See 8.25 for the definition of a convex set. This exercise should stretch your geometric intuition because this behavior cannot happen in finite dimensions.]

Give an example of a Banach space \(V,\) a normed vector space \(W,\) a bounded linear map \(T\) of \(V\) onto \(W\), and an open subset \(G\) of \(V\) such that \(T(G)\) is not an open subset of \(W\). [This exercise shows that the hypothesis in the Open Mapping Theorem that \(W\) is a Banach space cannot be relaxed to the hypothesis that \(W\) is a normed vector space. \(]\)
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