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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.

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 \( c_0 \).

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.