Inductive limit of $\mathbb R^n$s is Hausdorff and second countable?

Question

When dealing with infinite jet bundles, one can consider the topological vector space $\mathbb R^\infty$ obtained by taking the projective limit of the inverse system $(\mathbb R^n,\pi^n_m)$, where $\pi^n_m:\mathbb R^n\rightarrow\mathbb R^m$ ($m\le n$) is the projection onto the first $m$ factors. This space is topologized by taking the projective/initial topology on it, eg. the open sets are generated by preimages of open sets in $\mathbb R^n$ for various $n$s.

The dual construction is $\mathbb R^\infty_0$ is given by taking the inductive limit of the direct system $(\mathbb R^n,\imath^n_m)$, where $\imath^n_m:\mathbb R^m\rightarrow\mathbb R^n$ ($m\le n$) is the inclusion where the last $n-m$ factors are all zero. This space is topologized by taking the inductive/final topology where a set $U\subseteq\mathbb R^\infty_0$ is open if and only if $(\imath^\infty_n)^{-1}(U)$ is open for every $n$. Here $\imath^\infty_n:\mathbb R^n\rightarrow\mathbb R^\infty_0$ is the inclusion into the inductive limit.

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I have read about this in Saunders: The geometry of jet bundles, and he mentioned that $\mathbb R^\infty$ is Hausdorff and second-countable, but skips the proof and does not even mention it for $\mathbb R^\infty_0$. I have managed to prove that $\mathbb R^\infty$ is Hausdorff and second-countable, but it seems to me that the inductive topology of $\mathbb R^\infty_0$ is much more unfriendly than the projective topology of $\mathbb R^\infty$, and I have not managed to prove that $\mathbb R^\infty_0$ is Hausdorff and/or second-countable.

My question is about the topological properties of $\mathbb R^\infty_0$, is it Hausdorff and second-countable, and if so how can one prove it? I also do not know whether this notation is standard or not, since I have only seen this construction in Saunders (and in one other book which references Saunders there), so I cannot even search for papers or textbooks that would contain the information I want, as I do not know how $\mathbb R^\infty$ and $\mathbb R^\infty_0$ are called amongst mathematicians.

Answer 1

This space is often denoted by $\phi$ in functional analysis. It is separable, hausdorff but not second countable. Its bounded and compact sets resp. convergent sequences are the natural ones, i.e. the finite dimensional ones. It also has the specal property that you can take the inductive limit in the sense of topology, tvs'sor lcs's---they are all the same--and it is complete.

Answer 2

Among (infinite-dimensional) topologists the projective and inductive limits of the Euclidean spaces are denoted by $\mathbb R^\omega$ and $\mathbb R^\infty$, respectively. The topology of the (metrizable) space $\mathbb R^\omega$ was characterized by Henryk Torunczyk in 1980, and the topology of the (nonmetrizable) space $\mathbb R^\infty$ was characterized by Katsuro Sakai in 1984. Both characterizations can be found in Sakai's monograph "Topology of infinite-dimensional manifolds".

By the classical Anderson-Kadec Theorem, a locally convex space is homeomorphic to $\mathbb R^\omega$ if and only if it is infinite-dimensional and Polish (=separable and complete-metrizable). So, there are many non-isomorphic locally convex spaces, which are homeomorphic to $\mathbb R^\omega$. On the other hand, a locally convex space is topologically isomorphic to $\mathbb R^\infty$ if and only if it is homeomorphic to $\mathbb R^\infty$ (see, this paper).

Answer 3

A good reason why $\mathbb R^\infty_0$ is not metrizable (hence not first countable, a property that coincides with metrisability for Hausdorff TVS) is Baire’s theorem. You can follow this path: (1) A set is bounded in $\mathbb R^\infty_0$ iff is included in some $\mathbb R^n$ and bounded there. (2) A sequence converges to $x$ in $\mathbb R^\infty_0$ iff it is included in some $\mathbb R^n$ and converges to $x$ there. (3) A Cauchy sequence is always bounded (recall that a Cauchy sequence in a TVS is any $(x_n)_n$ such that for any nbd $U$ of $0$, there exists $n\in\mathbb N$ such that for all $p\ge n$ and $q\ge n$ There holds $x_p-x_q\in U$). As a conclusion, $\mathbb R^\infty_0$ is sequentially complete. If it were metrizable, it would be of second category by the Baire theorem, but this is not the case, because it is a countable union of closed sets with empty interior, namely the spaces $\mathbb R^n$.

Answer 4

Here is another (more elementary) way of proving the nonmetrizability of $\mathbb R_0^\infty$. Let $X=\mathbb R_0^\infty$, and let us denote the ``coordinates'' of an element $x\in X$ by $x^1$, $x^2$, etc. Thus $x$ is a sequence $(x^1,x^2,\ldots)$ of real numbers such that $x^n=0$ for all but finitely many $n$.

$\mathbf{(1)}$ For each $x\in X$, the map $\mathbb R\to X$, $\lambda\mapsto\lambda x$, is continuous. (This is immediate from the definition of the topology on $X$.)

$\mathbf{(2)}$ Assume, towards a contradiction, that $X$ is metrizable. Then for each sequence $(x_n)$ in $X$ there is a sequence $(\lambda_n)$ in $\mathbb R$, $\lambda_n>0$, such that $\lambda_n x_n$ converges to $0$. (Indeed, let $U_n$ denote the open ball of radius $1/n$, centered at $0$. By (1), for each $n$ there exists $\lambda_n>0$ such that $\lambda_n x_n\in U_n$. Clearly, $\lambda_n x_n\to 0$ as $n\to\infty$.)

$\mathbf{(3)}$ Let now $e_n=( \ldots ,0,1,0,\ldots )$ with $1$ in the $n$th slot, $0$ elsewhere. By (2), there is a sequence $(\lambda_n)$ in $\mathbb R$, $\lambda_n>0$, such that $\lambda_n e_n$ converges to $0$. Let now $$ U=\{ x\in X : |x^n|<\lambda_n\quad\text{for all}\; n\}. $$ Clearly, $U$ is a neighborhood of $0$ in $X$. On the other hand, $\lambda_n e_n\notin U$ for every $n$, and so $\lambda_n e_n$ does not converge to $0$. This is a contradiction.

In my view, the advantage of this proof is that it does not involve any specific notions from the theory of topological vector spaces (bounded sets, Cauchy sequences, completeness, etc.), so it is available to undergraduates learning elementary general topology.