# Subspaces of same finite codimension are isomorphic

I would like to show that two closed subspaces $Y$ and $Z$ of a normed space $X$ are isomorphic provided $\text{codim } Y = \text{codim } Z <\infty$.

I can show that $\text{codim}(Y\cap Z) <\infty$ but I don't know how to use that.

Can someone give me a hint.

$Y\cap Z$ has finite codimension in $Y$ and $Z$ too, so you can choose $y_1,\dots,y_n\in Y$ such that $Y=(Y\cap Z)\oplus \langle y_1,\dots,y_n\rangle$ (where $n:=\text{dim}\frac{Y}{Y\cap Z}$) and similarly there are $z_1,\dots,z_m\in Z$ such that $Z=(Y\cap Z)\oplus\langle z_1,\dots,z_m\rangle$.
Now we have $m=n$ (why?), so there is a linear isomorphism $\phi:\langle y_1,\dots,y_n\rangle\to\langle z_1,\dots,z_m\rangle$.
This extends to the isomorphism $\text{id}_{Y\cap Z}\oplus\phi:(Y\cap Z)\oplus \langle y_1,\dots,y_n\rangle\to(Y\cap Z)\oplus \langle z_1,\dots,z_m\rangle$, i.e. $\text{id}_{Y\cap Z}\oplus\phi:Y\to Z$ is an isomorphism, which is clearly continuous, as well as its inverse (why?).

A couple of things to add to Mizar’s excellent answer, which would show that it is valid for general topological vector spaces and not just for normed vector spaces:

• $Y \cap Z$ has finite co-dimension in $Y$ and $Z$ because of the 2nd Isomorphism Theorem: $$Y / (Y \cap Z) \cong (Y + Z) / Z \quad \text{and} \quad Z / (Y \cap Z) \cong (Y + Z) / Y.$$ As $Y + Z \subseteq X$, we have that $(Y + Z) / Z \subseteq X / Z$ and $(Y + Z) / Y \subseteq X / Y$; as $X / Y$ and $X / Z$ are finite-dimensional, we conclude that $Y \cap Z$ must have finite co-dimension in $Y$ and $Z$.

• The reason why $m = n$ is because $$1 + m = {\text{codim}_{X}}(Y) = {\text{codim}_{X}}(Z) = 1 + n.$$

• Let $V \stackrel{\text{df}}{=} \langle y_{1},\ldots,y_{m} \rangle$ and $W \stackrel{\text{df}}{=} \langle z_{1},\ldots,z_{m} \rangle$.

• Let $\phi: V \to W$ be any vector-space isomorphism. As $V$ and $W$ are finite-dimensional vector spaces, we have that $\phi$ and its inverse, $\phi^{\leftarrow}$, are automatically continuous.

• The mapping $\text{id}_{Y \cap Z} \oplus \phi: Y = (Y \cap Z) \oplus V \to (Y \cap Z) \oplus W = Z$ defined by $$\forall (x,v) \in (Y \cap Z) \times V: \quad (\text{id}_{Y \cap Z} \oplus \phi)(x \oplus v) \stackrel{\text{df}}{=} x \oplus \phi(v)$$ is shown to be continuous as follows. Suppose that $(x_{i} \oplus v_{i})_{i \in I}$ is a net in $Y$ that converges to $x \oplus v$. It is not obvious rightaway that $x_{i} \to x$ and $v_{i} \to v$ in $Y$. We cannot employ the Closed Graph Theorem here because we are dealing with a general topological vector space. However, we do have an abstract vector-space isomorphism $$Y / (Y \cap Z) \cong V$$ via the linear operator $T: (x \oplus v) + (Y \cap Z) \mapsto v$. Then $T$ is automatically continuous, being a vector-space isomorphism between finite-dimensional vector spaces. Hence, if $q$ denotes the continuous quotient mapping $Y \to Y / (Y \cap Z)$, then $T \circ q: Y \to V$ is continuous. Therefore, $v_{i} \to v$ in $V$ and so in $Y$, which means that $x_{i} \to x$ in $Y$ and so in $Y \cap Z$ and $Z$.

• In addition, $\phi(v_{i}) \to \phi(v)$ in $W$ and consequently in $Z$, so $x_{i} \oplus \phi(v_{i}) \to x \oplus \phi(v)$ in $Z$. This proves that $\text{id}_{Y \cap Z} \oplus \phi$ is continuous.

• By the same reasoning, $\text{id}_{Y \cap Z} \oplus \phi^{\leftarrow}: Z = (Y \cap Z) \oplus W \to (Y \cap Z) \oplus V = Y$ is continuous.

• As these two mappings are inverses, we conclude that $Y \cong Z$ as topological vector spaces.