Question

Show that \(\ell_{1}\) is not isomorphic to a subspace of \(c_{0}\). Hint: The dual of \(\ell_{1}\) is nonseparable.

Short answer

\(\ell_{1}\) is not isomorphic to a subspace of \(\c_{0}\) because \(\ell_{1}\) is nonseparable while \(\c_{0}\) and its subspaces are separable.

Step-by-step solution

- Understanding \(\ell_{1}\) and \(\c_{0}\)

The space \(\ell_{1}\) consists of all absolutely summable sequences \((x_{n})\) such that \(\[ \sum_{n=1}^{\backslashinfty} |x_{n}| < \backslashinfty \]\). The space \(\c_{0}\) consists of all sequences \((x_{n})\) that converge to 0. Both are infinite-dimensional vector spaces, but their properties differ significantly.

- Use the Hint on Duality

Recall the hint stating that the dual of \(\ell_{1}\), denoted \(\ell_{\backslashinfty}\), is nonseparable. This means there does not exist a countable dense subset in \(\ell_{\backslashinfty}\) with respect to the weak* topology.

- Check Separability of \(\c_{0}\)

Note that \(\c_{0}\) is a separable space. This implies any subspace of \(\c_{0}\) is also separable. A separable space has a countable dense subset.

- Contradiction in Isomorphism

If \(\ell_{1}\) were isomorphic to a subspace of \(\c_{0}\), then \(\ell_{1}\) would also have to be separable, because all subspaces of a separable space are separable. However, since \(\ell_{1}\) is not separable (as its dual \(\ell_{\backslashinfty}\) is nonseparable), this leads to a contradiction.

- Conclusion

We conclude that \(\ell_{1}\) cannot be isomorphic to a subspace of \(\c_{0}\) due to the difference in separability properties.

Key concepts

isomorphism

An isomorphism between two vector spaces is essentially a mapping that preserves the structure and properties of those spaces. This mapping—called an **isomorphism**—must be both one-to-one (injective) and onto (surjective). Simply put, an isomorphism between vector spaces \(V\) and \(W\) maps elements from \(V\) to \(W\) in such a way that all the operations in \(V\) have their counterparts in \(W\), and vice versa. Here are some key points to remember about isomorphisms:

• An isomorphism maintains vector addition and scalar multiplication.
• If \(V\) and \(W\) are isomorphic, they are essentially the same in terms of vector space properties.
• Isomorphisms are special because they indicate a perfect 'match' between spaces, despite possible differences in their presentation or nature.

When dealing with spaces like \( \ell_1 \) and \( c_0 \), identifying an isomorphism means finding a bijective linear map that respects the space structures. Since we've seen that \( \ell_1 \) cannot be isomorphic to any subspace of \( c_0 \), this indicates deep-rooted structural differences, especially in terms of separability, which we'll explore next.

dual space

The concept of a **dual space** is fundamental in functional analysis. For any vector space \(V\), the dual space, denoted as \(V^*\), consists of all linear functionals on \(V\). A linear functional is essentially a linear mapping from \(V\) to the field over which \(V\) is defined, usually the field of real or complex numbers.

• The dual space captures 'how' functions can act on the elements of \(V\).
• For example, the dual of \( \ell_1 \) is \( \ell_{\backslashinfty} \), which contains all bounded sequences.
• The properties of a space's dual can sometimes be starkly different from the original space itself.

In the specific exercise given, we needed to recognize that \( \ell_{\backslashinfty} \) (the dual space of \( \ell_1 \)) is nonseparable. This concept of separability—where a space has a countable dense subset—becomes crucial in the next part of the argument:
Since \( \ell_1 \) maps through its dual \( \ell_{\backslashinfty} \), and \( \ell_{\backslashinfty} \) lacks a countable dense subset, understanding dual spaces helps us reason why \( \ell_1 \) cannot fit into any separable subspace like those in \( c_0 \).

separability

A vector space is called **separable** if it has a countable dense subset. This means there is a countable set of points in the space such that any point in the space can be approximated as closely as desired by points from this countable set.

• Separable spaces are important in analysis because they allow the usage of sequences, a common tool in mathematical proofs.
• Both metric spaces and vector spaces can exhibit separability.
• For instance, \( c_0 \), the space of sequences converging to zero, is separable.

Returning to our specific exercise, \( c_0 \) being separable implies that any subspace of \( c_0 \) is also separable. But, the problem arises when considering \( \ell_1 \), a space whose dual \( \ell_{\backslashinfty} \) is known to be nonseparable.
If \( \ell_1 \) were a subspace of \( c_0 \), it would inherit separability from \( c_0 \). However, since \( \ell_1' \) (the dual) is nonseparable, this is impossible. Thus, understanding separability and its implications allows us to see why \( \ell_1 \) cannot be a subspace of \( c_0 \), leading to a contradiction and ultimately proving the exercise.