Limit of Euclidean Spaces

Question

Consider the projective system of topological spaces $(\mathbb{R}^d,\pi_n^m)$ where $\pi_n^m(x_1,\dots,x_m)\mapsto (x_1,\dots,x_n)$ where $n\leq m$. Is the projective limit $$ \projlim_{\iota_n^m} \mathbb{R}^d \cong \prod_{d \in \mathbb{N}} \mathbb{R}? $$

Dually, consider $\iota_n^m:(x_1,\dots,x_n)\mapsto (x_1,\dots,x_n,0,\dots,0)$ mapping $\mathbb{R}^n$ into $\mathbb{R}^m$; then $(\mathbb{d},\iota_n^m)$ defines an inductive system of topological spaces. Then is $$ \injlim_{\iota_n^m} \mathbb{R}^d \cong \ell^2? $$

Answer 1

Let $\mathbb{R}^\infty = \prod_{n \in \mathbb{N}} \mathbb{R}$ denote the countable infinite product of copies of $\mathbb R$. Let $\pi_d : \mathbb{R}^\infty \to \mathbb R^d$ denote the projection onto the first $d$ coordinates (which is continuous). We shall verify that this system has the universal property of the inverse limit.

So let $f_d : X \to \mathbb{R}^d$ be a family of maps such that $f_d = \pi^m_d f_m$ for $m \ge d$.

For $n \le m \le \infty$ let $p^m_n : \mathbb{R}^m \to \mathbb R$ denote the projection onto the $n$-th coordinate. Define $$f : X \to \mathbb{R}^\infty, f(x) = (p^n_n f_n(x))_{n \in \mathbb{N}} .$$ Then $\pi_d f = f_d$. To see this, it suffices to show that $p^d_n \pi_d f = p^d_n f_d$ for all $n \le d$. But we have $p^d_n \pi_d = p_n^\infty$ so that $p^d_n \pi_d f = p^n_n f_n = p^n_n \pi^d_n f_d = p^d_n f_d$.

Next let $f' : X \to \mathbb{R}^\infty$ be a map such that $\pi_d f' = f_d$ for all $d$. Then $p^\infty_n f' = p^n_n \pi_n f' = p^n_n f_n = p^\infty_n f$ for all $n$, i.e. $f' = f$.

Concerning your second question: It am not sure whether it is true. The space $\ell^2$ is defined as the subset $\ell^2 = \{ (x_n) \in \mathbb{R}^\infty \mid \sum_{n=1}^\infty x_n^2 \text{ is convergent} \}$. This is a normed linear space with the norm $\lVert (x_n) \rVert_2 = \sqrt{\sum_{n=1}^\infty x_n^2}$. This norm induces the topology on $\ell^2$.

The direct limit of your direct system can be determined as follows. Let $\bigoplus_{n=1}^\infty \mathbb R = \{ (x_n) \in \mathbb{R}^\infty \mid x_n = 0 \text{ for almost all } n \}$. Endow it with the finite topology which means that $U \subset \bigoplus_{n=1}^\infty \mathbb R$ is open iff for each finite-dimensional linear subspace $E \subset \bigoplus_{n=1}^\infty \mathbb R$ the set $U \cap E$ is open in $E$. Note that finite-dimensional linear spaces $E$ have a canonical topology (this is the unique Hausdorff topology making $E$ a topological vector space; it is induced by any norm on $E$).

Let $\iota_d : \mathbb R^d \to \bigoplus_{n=1}^\infty \mathbb R, \iota_d(x_1,\ldots,x_d) = (x_1,\ldots,x_d,0,0,\ldots)$, and $E_d = \iota_d(\mathbb R^d)$. Then $E_d \subset E_{d+1}$ and $\bigcup_{n=1}^\infty E_n = \bigoplus_{n=1}^\infty \mathbb R$. It is easy to see that $U \subset \bigoplus_{n=1}^\infty \mathbb R$ is open iff for each $n$ the set $U \cap E_n$ is open in $E_n$. Moreover, the $\iota_d$ are continuous.

We claim that the system of the $\iota_d$ is a direct limit of your direct system. This is equivalent to showing that the direct limit of the system $(E_d,i^m_n)$ with bondings $i^m_n : E_m \hookrightarrow E_n$ is given by the inclusions $i_d : E_d \hookrightarrow \bigoplus_{n=1}^\infty \mathbb R$. But this is an easy job which I leave to you. The crucial point is that a map $g : \bigoplus_{n=1}^\infty \mathbb R \to Y$ is continuous iff all $g i_d$ are continuous.

More interesting are:

1. $\bigoplus_{n=1}^\infty \mathbb R$ is a subset of $\ell^2$ and can be endowed with the norm-topology. However, the norm topology does not agree with the finite topology. In fact, define $g : \bigoplus_{n=1}^\infty \mathbb R \to \mathbb R, g((x_n)) = \sum_{n=1}^\infty nx_n$. This is a linear map which is trivially contiunuos with respect to the finite topology. It is not contiunuos with respect to the norm topology. To see this, consider $\xi_k \in \bigoplus_{n=1}^\infty \mathbb R$ having an entry $1/k$ in the $k$-th coordinate and all other coordinates $0$. Then $\xi_k \to 0$ in the norm topology, but $g(\xi_k) = 1$ which does not converge to $0 = g(0)$.

2. Nevertheless it could be that $\bigoplus_{n=1}^\infty \mathbb R$ with the finite topology is homeomorphic to $\ell^2$. I do not know whether this is true, but I doubt it. Anyway, if it were true, then we would not have obviuous maps $\mathbb R^d \to \ell^2$ producing a direct limit.