Isomorphism from Cartesian Space to Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Homeomorphism

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a Hausdorff topological vector space over $\GF$.

Let $n \in \N$.

Let $Y$ be a subspace of $X$ with dimension $n$.

Let $f : \GF^n \to Y$ be a vector space isomorphism.


Then $f$ is a homeomorphism.


Proof

First, we assure ourselves that such a vector space isomorphism $f : \GF^n \to Y$ exists from Same Dimensional Vector Spaces are Isomorphic.

From Linear Transformation from Cartesian Space on Hausdorff Topological Field to Topological Vector Space is Continuous, $f$ is continuous.

It suffices to show that $f^{-1}$ is continuous.

Let $S$ be the unit sphere in $\GF^n$.

Let $B$ be the unit ball in $\GF^n$.

From Heine-Borel Theorem: Normed Vector Space, we have that $S$ is compact.

Let:

$K = f \sqbrk S$

From Continuous Image of Compact Space is Compact, $K$ is compact.

From Compact Subspace of Hausdorff Space is Closed, $K$ is closed in $X$.

Note that $\ker f = \set { {\mathbf 0}_X}$, since $f$ is injective, so we have $\mathbf 0_X \not \in K$.

From Characterization of Closure by Open Sets, there exists an open neighborhood $U$ of $\mathbf 0_X$ such that:

$U \cap K = \O$

From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists a balanced open neighborhood of $\mathbf 0_X$ such that $V \subseteq U$.

From the definition of the subspace topology we then obtain that $V \cap Y$ is a balanced open neighborhood of $\mathbf 0_X$ in $Y$ such that:

$\paren {V \cap Y} \cap K = \O$

Now set:

$E = f^{-1} \sqbrk {V \cap Y} = f^{-1} \sqbrk V$

This is non-empty, since $0 \in E$.

Further, we show that $E$ and $S$ are disjoint sets.

If $x \in E \cap S$ we would have:

$\map f x \in K$

and:

$\map f x \in E$

But since $E \cap S = \O$, this is impossible.

Since $f$ is linear, $f^{-1}$ is linear by Inverse of Linear Transformation is Linear Transformation.

Since $V$ is balanced, we have that $E = f^{-1} \sqbrk V$ is balanced from Image of Balanced Set under Linear Transformation is Balanced.

From Balanced Set in Topological Vector Space is Connected, $E$ is connected.

We argue that we must have $E \subseteq B$.

Let $D$ be the closed unit ball in $\GF^n$.

We can then write:

$\GF^n = B \cup S \cup \paren {\GF^n \setminus D}$

So, we may write:

$E = \paren {B \cap E} \cup \paren {E \cap \paren {\GF^n \setminus D} }$

from Intersection Distributes over Union, since $E \cap S = \O$.

Since $\GF^n \setminus D$ is open, if we had:

$E \cap \paren {\GF^n \setminus D} \ne \O$

then $E$ would be disconnected as the union of disjoint open sets.

We have determined that $E$ is connected, so we have:

$E \cap \paren {\GF^n \setminus D} = \O$

and so:

$B \cap E = E$

giving:

$f^{-1} \sqbrk V = E \subseteq B$

From Characterization of von Neumann-Boundedness in Normed Vector Space, we have that $B$ is von Neumann bounded.

So $f^{-1}$ maps an open neighborhood of $\mathbf 0_X$ into a von Neumann bounded set.

It follows from Characterization of Continuous Linear Functionals on Topological Vector Space that $f^{-1}$ is continuous.

$\blacksquare$


Sources

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