Question

Let \(X\) be a Banach space and \(G\) be a subspace of \(X\) that is a \(G_{\delta}\) set in \(X\). Show that \(G\) is closed in \(X\). Hint: Let \(S=\bar{G} \backslash G\), where \(\bar{G}\) denotes the closure of \(G .\) Since \(G\) is a \(G_{\delta}\) set in \(X, G\) is \(G_{\delta}\) in \(\bar{G}\). Hence \(G=\bigcap G_{n}\), where \(G_{n}\) are open subsets in \(\bar{G}\). Therefore, \(S=\bigcup\left(\bar{G} \backslash G_{n}\right)\) and each \(G_{n}\) is dense in \(\bar{G}\) because it contains \(G .\) Thus, each \(\bar{G} \backslash G_{n}\) is nowhere dense in \(\bar{G}\), so \(S\) is of first category in \(\bar{G}\). We will show that \(S\) is an empty set. If \(x_{0} \in S\), consider the set \(G^{*}=\left\\{x_{0}+x ; x \in G\right\\}\). Note that \(G^{*} \subset S\). Indeed, any point \(x_{0}+x, x \in G\), is in \(\bar{G}\) since \(\bar{G}\) is a linear set. If for some \(x \in G\) we have \(x_{0}+x \in G\), then by linearity of \(G\) we have \(x_{0} \in G_{1}\) a contradiction. Therefore, \(G^{*} \subset S\) and \(G^{*}\) is of first category in \(\bar{G} .\) Thus, the shift \(G\) of \(G^{*}\) is also of first category in \(\bar{G}\). Hence, the whole \(\bar{G}\) as a union of \(G\) and \(S\), which are both of first category in \(\bar{G}\), is of first category in itself. This is a contradiction, since \(\bar{G}\) is a complete metric space.

Short answer

Since \(S = \bar{G} \backslash G\) must be empty to avoid contradictions, \(G = \bar{G}\). Therefore, \(G\) is closed in \(X\).

Step-by-step solution

- Understand the problem

We need to demonstrate that if a subspace \(G\) of a Banach space \(X\) is a \(G_{\boxed{}}\text{-set}\) in \(X\), then \(G\) is closed in \(X\). A \(G_{\boxed{}}\text{-set}\) is a countable intersection of open sets.

- Define the closure \(\bar{G}\) and set \(S\)

Let \(\bar{G}\) denote the closure of \(G\) in \(X\). Define \(S = \bar{G} \backslash G\). We want to show that \(S = \boxed{0}\), i.e., \(S\) is empty.

- Use the fact that \(G\) is a \(G_{\boxed{}}\text{-set}\)

Since \(G\) is a \(G_{\boxed{}}\text{-set}\) in \(X\), there exist open sets \(G_n\) such that \(G = \bigcap G_n\), where each \(G_n\) is open in \(\bar{G}\).

- Express \(S\) in terms of \(G_n\)

Note that \(S = \bigcup (\bar{G} \backslash G_n)\). Each \(G_n\) is dense in \(\bar{G}\) since \(G_n\) contains \(G\). Therefore, \(\bar{G} \backslash G_n\) is nowhere dense in \(\bar{G}\).

- Show \(S\) is of first category

Since \(\bar{G} \backslash G_n\) is nowhere dense in \(\bar{G}\) and \(S = \bigcup (\bar{G} \backslash G_n)\), it follows that \(S\) is of first category in \(\bar{G}\).

- Assume for contradiction that \(S\) is non-empty

Assume for contradiction that there exists \(x_0 \boxed{\text{in}} S\). Define \(G^* = \boxed{x_0 + G} = \boxed{\boxed{}}x_0{} + G = \boxed{\boxed{x+x_0}}{}; x \boxed{\boxed{\text{in}} G\).}

- Show \(G^* \subseteq S\)

Any \(x_0 + x\) with \(x \boxed{\text{in}} G\) must be in \(\bar{G}\) since \(\bar{G}\) is closed under addition. Since \(G\) is linear, \(x_0 + x\) cannot be in \(G\). Therefore, \(G^* \subseteq S\).

- Conclude that \(G^*\) is of first category

Since \(G^*\) is a shift of \(G\), it is also of first category in \(\bar{G}\).

- Derive the contradiction

The union of \(G\) and \(S\) forms \(\bar{G}\), which implies that \(\bar{G}\) would be of first category. This contradicts the fact that \(\bar{G}\) is a complete metric space.

- Conclude

Thus, the assumption that \(S\) is non-empty is false. Hence, \(S\) must be empty, and \(G = \bar{G}\). Therefore, \(G\) is closed in \(X\).

Key concepts

G-delta set

In topology, a G-delta set (or G_δ set) refers to a set that can be expressed as a countable intersection of open sets. Suppose you have a sequence of open sets \(\{G_n\}\). Then, a G-delta set is defined as: \(\bigcap_{n=1}^{\text{∞}} G_n\). This property is significant because G-delta sets preserve certain 'nice' properties of open sets even though they might be more complicated in structure.
In the context of the provided exercise, if G is a subspace of a Banach space X and is also a G-delta set, this property becomes key to proving that G is closed. The G-delta property allows us to represent G as the intersection of a sequence of open sets, thereby making it easier to analyze its structure and relate it to the properties of closed sets and their complements.

closed subspaces

A closed subspace of a topological space is a subset that contains all its limit points. This means, in a topological space X, a subspace G is closed if every limit point of G is also in G. In other words, if a sequence of points in G converges to a point, that point must also lie in G. In metric or normed spaces like Banach spaces, being closed also means that G contains all its boundary points.
The importance of closed subspaces in the context of this exercise is that we are tasked to show that the G-delta set G within a Banach space G is also closed. The concept of closures in metric spaces helps us understand that the complement of G within its closure forms a set of the first category—meaning it is 'small' in a certain sense—and demonstrating this leads us to prove G's closedness.

complete metric space

A complete metric space is one where every Cauchy sequence (a sequence where the elements get arbitrarily close to each other as the sequence progresses) converges to a point within the space. For instance, the set of real numbers with the usual distance metric is a complete metric space because any Cauchy sequence of real numbers converges to a real number.
For Banach spaces (which are complete normed vector spaces), this property ensures that limit processes behave well within the space. In the provided exercise, the completeness of the space \(\bar{G}\) plays a crucial role. When demonstrating the structure of the complement of G within its closure and indicating that the space itself cannot be of the first category, we use completeness to reach a contradiction. This allows us to conclude that G must be closed.