Topological Vector Space: $\dim Z\text{ finite}\implies Z\text{ closed}$

Question

Let $V$ be a Hausdorff topological vector space and $Z$ a linear subspace: $Z\leq X$

Is there a neat way to prove that: $$\dim Z\text{ finite}\implies Z\text{ closed}$$

Answer 1

A finite-dimensional Hausdorff topological vector space (over $\mathbb{C}$ or $\mathbb{R}$) is locally compact, hence complete.$^1$

A complete subspace of a Hausdorff space is closed.

Normed vector spaces are Hausdorff.

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$^1$ Namely, the isomorphism $\varphi_B \colon \mathbb{K}^n \to E; x \mapsto \sum_{k=1}^n x_k\cdot b_k$ induced by a basis $B = \{b_k : 1 \leqslant k \leqslant n\}$ of the $n$-dimensional vector space $E$ over $\mathbb{K}$ is a topological isomorphism. Since $\mathbb{K}^n$ is locally compact (a consequence of the Heine-Borel theorem for example), consequently so is $E$.

That a locally compact (Hausdorff) topological vector space is complete follows for example from Riesz' theorem that a Hausdorff TVS is locally compact if and only if it is finite-dimensional, and the known fact that $\mathbb{K}^n$ is complete.

It can also be proved without that. In the metric case we can observe that Cauchy sequences are bounded, and hence the underlying set of a Cauchy sequence is relatively compact. Therefore every Cauchy sequence has convergent subsequences, but a Cauchy sequence converges to all its accumulation points.

If we don't assume the space to be metrisable (if it is locally compact, it is metrisable, but we don't need to use that fact), we take a compact neighbourhood $C$ of $0$ and observe that every Cauchy filter contains a set that is small of order $C$, and therefore relatively compact. From that we deduce that every Cauchy filter has an accumulation point, and again, Cauchy filters converge to all their accumulation points.

Answer 2

If you just want to stick to norms, choose a basis $\{ z_{1},\cdots,z_{n}\}$ for $Z$ and define a map $L : \mathbb{C}^{n}\rightarrow Z$ by $L(\alpha_{1},\cdots,\alpha_{n})=\sum_{j=1}^{n}\alpha_{j}z_{j}$. This map $L$ is continuous because $$ \|L(\alpha_{1},\cdots,\alpha_{n})\|_{X} \le \sum_{j}|\alpha_{j}|\|z_{j}\| \le \left(\sum_{j}\|z_{j}\|_{X}^{2}\right)^{1/2}\left(\sum_{j}|\alpha_{j}|^{2}\right)^{1/2} = C\|(\alpha_{1},\cdots,\alpha_{n})\|_{\mathbb{C}^{N}}. $$ So $L$ is continuous because $\|Lx\|_{X}\le C\|x\|_{\mathbb{C}^{n}}$ for all $x\in\mathbb{C}^{n}$, where $C$ is the obvious constant.

Next we show that $L^{-1}$ is also continuous. To do this, let $S$ be the unit sphere in $\mathbb{C}^{N}$ given by $S=\{ x\in \mathbb{C}^{n} : \|x\|_{\mathbb{C}^{n}}=1\}$. $S$ is compact in $\mathbb{C}^{n}$ from classical results. The function $l(x)=\|Lx\|_{X}$ is continuous on $\mathbb{C}^{n}$ and does not vanish on $S$. Therefore $l$ has a minimum on $S$ which is non-zero. That is $\|Lx\|_{X}\ge \rho >0$ for some constant $\rho$ and for all $x \in S$. Equivalently, $\|Lx\|_{X}\ge\rho\|x\|_{\mathbb{C}^{n}}$ for all $x \in \mathbb{C}^{n}$. So $L : \mathbb{C}^{n}\rightarrow Z$ is a topological isomorphism because $L$ and $L^{-1}$ are bounded.

It follows almost immediately that $Z$ is closed in $X$. If $\{ y_{n} \}_{n=1}^{\infty}\subseteq Z$ converges in $X$ to some $y_{0}\in X$, then $\{ y_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence in $Z$, which implies that $\{ L^{-1}y_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{C}^{n}$ ($L^{-1}$ is continuous) and, hence, converges to some $x_{0}\in \mathbb{C}^{n}$. It then follows that $\{ y_{n}\}_{n=1}^{\infty}$ must converge to $Lx_{0}$ ($L$ is continuous.) And that forces $\overline{Z}\subseteq Z$.

NOTE: This is the standard argument given to show that all norms on finite-dimensional normed linear spaces are equivalent.

Answer 3

We must assume that $V$ is Hausdorff; otherwise this is false (let $V$ have dimension at least 1 and equip it with the indiscrete topology, then the zero subspace is not closed).

You may find a complete proof, with no further assumptions on the topology of $V$, at Theorem 1 in these notes by Gabriel Nagy. (Basic definitions are in this file; note in particular that for Nagy, "linear topology" does not assume Hausdorff but "topological vector space" does. If you should wish to refer to other sections of the notes, the index is here.) His proof is rather clever; using induction on $n$, it proves the stronger statement "For each $n$, there is a unique Hausdorff TVS topology on $\mathbb{K}^n$, and every $n$-dimensional subspace of $V$ is closed".

Answer 4

Proof:

Let $Z$ be a finite dimensional subspace of a Hausdorff topological vector space $X$ and $E$ an Euclidean vector space of the same dimension.

Since both dimensions agree there is a linear injection $L:E\to V$ onto $Z$.

From here we know that $L$ is continuous and that $U$ constructed in there is a balanced and open(!) neighborhood of zero in the topology of $X$ that doesn't intersect $LS$ for the unit sphere $S$ in $E$. This time the crucial observation will be $Z\cap U\subseteq LD$ with the closed unit ball $D$ ('D' for disk).

Lemma: $Z\cap U\subseteq LD$
By construction $U$ is disjoint from $LS$ so also $Z\cap U$. Since $L$ is injective $L^{-1}(Z\cap U)$ is disjoint from $S$. But $L^{-1}(Z\cap U)$ is also balanced by linearity of $L$. Therefore $L^{-1}(Z\cap U)$ is contained in the open unit ball $B$ and moreover in $D$ and so $Z\cap U\subseteq LD$ since $Z\cap U$ was in the image of $L$.

Now, let $p\in \overline Z$. Since $U$ is a neighborhood of zero it is absorbing that is $p\in\delta U$ for some $\delta>0$. But $U$ was especially open so $\delta U$ is a neighborhood of $p$. Thus $p$ is in the closure of $Z\cap\delta U$. On the same time since $\delta D$ in $E$ is compact $L\delta D$ is compact in $X$ as $L$ is continuous. But as $X$ is Hausdorff $L\delta D$ is closed in $X$. Thus $\overline{Z\cap\delta U}\subseteq\overline{L\delta D}=L\delta D\subseteq Z$. So $p$ was already in $Z$.

Concluding that $Z$ is closed in $X$. c.q.f.d.