Let's break down your two questions clearly, using standard linear algebra conventions that apply to both finite and infinite-dimensional spaces.

1. Does the codimension definition hold when both $Y$ and $X/Y$ are infinite-dimensional? #

Absolutely—this definition remains fully meaningful, but we need to clarify what "dimension" means for infinite spaces. For any linear space (finite or infinite), dimension is defined as the cardinality (size) of its Hamel basis (a maximal linearly independent set).

Even when $Y$ and $X/Y$ are both infinite-dimensional, their Hamel bases have well-defined cardinalities (we use infinite cardinal numbers here, which follow consistent arithmetic rules). So $\text{codim}(Y) = \dim(X/Y)$ simply defines codimension as the cardinality of $X/Y$'s Hamel basis—a concrete, valid concept in this context.

For a concrete example:

• Let $X$ be the space of all real-valued sequences (its Hamel basis has uncountable cardinality).
• Let $Y$ be the subspace of sequences with only finitely many non-zero terms (its Hamel basis is countable).
• The quotient space $X/Y$ has an uncountable Hamel basis, so $\text{codim}(Y)$ is an uncountable cardinal—this is a meaningful value under the definition.

2. Why does $\dim(X/Y) + \dim(Y) = \dim(X)$ hold for both finite and infinite-dimensional spaces? #

The core reasoning relies on constructing a Hamel basis for $X$ using bases of $Y$ and $X/Y$:

• Start with a Hamel basis $B_Y$ for $Y$.
• Since $B_Y$ is linearly independent in $X$, we can extend it to a full Hamel basis for $X$: $B_X = B_Y \cup B'$, where $B'$ is a set of vectors in $X$ that are linearly independent modulo $Y$.

Now look at the set ${v + Y \mid v \in B'}$ in $X/Y$:

• This set is linearly independent: If a linear combination of these equivalence classes equals the zero element of $X/Y$ (i.e., $Y$), the corresponding linear combination of vectors in $B'$ lies in $Y$. But since $B_Y$ and $B'$ are linearly independent in $X$, all coefficients must be zero.
• This set spans $X/Y$: Any element of $X/Y$ is $x + Y$ for some $x \in X$. Write $x$ as a linear combination of vectors from $B_Y$ and $B'$—the $B_Y$ terms vanish modulo $Y$, so $x + Y$ is a linear combination of elements from ${v + Y \mid v \in B'}$.

This makes ${v + Y \mid v \in B'}$ a Hamel basis for $X/Y$, so $\dim(X/Y) = |B'|$. We know $\dim(Y) = |B_Y|$ and $\dim(X) = |B_Y \cup B'|$. Since $B_Y$ and $B'$ are disjoint, the cardinality of their union is the sum of their cardinalities:
$$|B_Y \cup B'| = |B_Y| + |B'|$$

For finite-dimensional spaces, this is ordinary integer addition. For infinite-dimensional spaces, cardinal arithmetic tells us that adding two cardinals (where at least one is infinite) gives the larger of the two cardinals—this still satisfies the equality $\dim(X/Y) + \dim(Y) = \dim(X)$ when we interpret "addition" as cardinal addition.

内容的提问来源于stack exchange，提问作者A Slow Learner