Subspaces of same finite codimension are isomorphic

Question

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.

Answer 1

$Y\cap Z$ 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?).

Answer 2

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.