
  constant sequences in$ X_{\infty}$, the sequences whose
  terms coincide from index $n$ upward. Thus, with moderate abuse of notation,
  we extend the map $R_N |X_{N + 1}$ canonically to $R_N : X_{\infty}
  \rightarrow X_N$, so that $R_N (x_{1,} x_2, \ldots x_N, x_{N + 1}, \ldots.)
  = (x_{1,} x_2, \ldots x_N, x_N, \ldots)$, and note $R_N = R_N R_{N + 1}.$
  
  By definition a point $g \in G$ is a subspace $g \subset X_{\infty}$ so
  that distinct points $\{ x, y \} \subset g$ have $q_n$ equivalent
  coordinates for each $n$,  and a point $g_n \in G_n$ is a subspace $g_n \in
  X_n$. Our starting assumptions ensure we have a well defined function $G_n
  \hookrightarrow G$ sending $g_n \in G_n$ to $(\ldots.r_{n - 1} (g_n), g_n,
  j_n (g_n)_{}, j_{n + 1} j_n (g_n) \ldots) \in G$. This is a topological
  embedding, and henceforth we conflate $G_n$ with the corresponding subspace
  of $G$. The map $q_n R_n : X_{\infty} \rightarrow G_n$ is constant on sets
  of the form $q^{- 1} (g)$ and thus there is a unique induced map $\Pi_n : G
  \rightarrow G_n$ such that $\Pi_n = \Pi_n \Pi_{n + 1}.$ The maps $\{ \Pi_n
  \}$ determine a continuous injection $\phi : G \rightarrow \lim_{\leftarrow}
  G_n$.
  
  If $\phi$ were a topological embedding then it would follow easily that
  ${dimG} = 0$, since $\phi$ embeds $G$into the 0 dimensional metric
  space $G_1 \times G_2 \ldots$However in general $\phi$ is NOT a topological
  embedding {\cite{FA0}}. \
  
  Note each space $G_n$ is discrete, since topological quotients of discrete
  spaces are discrete. Moreover $G$is a quotient of a metrizable space and
  hence $G$is sequential. Thus axiom \ref{retracts} will hold provided we can
  show pointwise convergence $\Pi_n \rightarrow {id} |G.$ The latter
  claim holds by definition, shown as follows. Given $g \in G$ and an open $U
  \subset G$ so that $g \in U$, lift $g$ to some $(x_1, x_2, \ldots) \in q^{-
  1} (g) \subset X_{\infty}$ and obtain a basic open $V \subset q^{- 1} (U)
  \subset X_{\infty}$ with $(x_1, x_2, \ldots) \in V$ so that $V = \{ x_1 \}
  \times \{ x_2 \} \ldots \times \{ x_N \} \times X_{N + 1} \ldots.$Now given
  $n \geq N$ to check $\Pi_n (g) \in U$ it suffices to check, since $q :
  X_{\infty} \rightarrow G$ is a quotient map, that some lift of $\Pi_n (g)
  \in U$. Our special point $(x_1, x_2, \ldots) \in V$ suffices. Thus axiom
  \ref{retracts} holds.
  
  The space $\lim_{\leftarrow} X_n = X_{\infty}$ is a topological inverse
  limit of discrete spaces under retraction bonding maps $R_n |X_{n + 1}.$ We
  have well ordered each set $X_n $ so that $R_n (x_{n + 1}) \leq x_n
  $for $x_{n + 1} \in X_{n + 1}.$ Thus with the induced lexigraphic order on
  $X_{\infty}$ we have $R_n (x) \leq x$for $x \in X_{\infty}.$
  
  The space $X_{\infty} $ has the topology of pointwise convergence, and
  thus, since $X_n$ is a discrete space, a sequence $\{ s_n \} \subset
  X_{\infty}$ converges iff $\Pi_N (s_n)$ is eventually constant for each $N$.
  Thus, if the strictly increasing sequence $s_1 < s_2 \ldots \subset
  X_{\infty}$\quad diverges, then every subsequence of $\{ s_n \}$ diverges,
  since $\{ \Pi_N (s_n) \}$ is nondecreasing for each $N.$ Conversely, if the
  increasing sequence $\{ s_n \} \subset X_{\infty}$ converges, then, since
  $\{ \Pi_N (s_n) \}$ is nondecreasing and eventually constant for each $N,$
  $\lim \{ s_n \} = \sup \{ s_n \}$.
  
  Since each set $X_n$ is well ordered, and since no well ordered set admits
  a strictly decreasing sequence with infinitely many terms, each strictly
  decreasing sequence $s_1 > s_2 \ldots. \subset X_{\infty}$ converges.
  Consequently if $B \subset X_{\infty}$ is nonempty and closed, then $\min
  (B)$ exists. To see why, let $b_1 = \min \Pi_1 (B)$. Let $b_2 = \min \Pi_2
  \Pi_1^{- 1} (b_1).$ Let $b_3 = \min \Pi_3 \Pi_2^{- 1} (b_2)$ and so on.
  Note $b_1 \geq b_2 \ldots$ and let $b = \lim \{ b_n \}$. Thus $b \in B$
  since $B$ is closed.Note $b \leq a$ for each $a \in B$
  and thus $b = \min (B)$.
  
  Note $G$ is $T_2 $ by Remark \ref{gt2} and in particular $G$ is $T_1$.
  Thus point preimages are closed under the map $q : X_{\infty} \rightarrow
  G$. Define $\sigma : G \rightarrow X_{\infty}$ as $\sigma (g) = \min (q^{-
  1} (g))$. Since $\sigma$is one to one and since subsets of linearly ordered
  sets are linearly ordered, we obtain a linear order on $G$ defined so that
  $g < h$ iff \ $\sigma (g) < \sigma (h).$ In particular the sequence $\{ g_n
  \}$ is strictly increasing in $G$ iff $\{ \sigma (g_n) \}$ is strictly
  increasing in $X_{\infty}$. In the process of checking axiom \ref{lin} we
  will show $\sigma$ is left continuous but not right continuous.
  
  First we observe some basic properties of our definition of the linear order
  on the set $X_{\infty}$, conflating $X_N$ with the subspace of $X_{\infty}$,
  all of whose terms are constant from index $N$ and above. If $x = (x_1, x_2,
  \ldots.) \in X_{\infty}$ then $x_n \leq x_{n + 1} \leq x,$and
  $x_n \rightarrow x,$ and $x = \sup \{ x_n \}.$
  
  Next we observe a basic property of $\sigma$. If $(y_1, y_2, \ldots.) = y
  < z = \sigma (g) = (z_1, z_2, \ldots) \in X_{\infty}$ and if $N$ is minimal
  so that $y_N \neq z_N$, then $y_N < z_N$, $\min (q_N (y_N)) < z_N = \min
  (q_N (z_N))$ and hence $q_N (y_N) < q_N  (z_N).$ \
  
  To check axiom \ref{lin}, if $A \subset G$ is closed, then $q^{- 1} (A) =
  B$ is closed in $X_{\infty}$ and thus ${minA} = q ({minB})$ and in
  particular $A$ has a minimal point. Suppose $\{ g_n \}$ is a strictly
  increasing sequence in $G$.
  
  Case 1. Suppose the corresponding increasing sequence $\{ \sigma (g_n) \}$
  converges to some $x = (x_{1,} x_2,