Theorem 1 (Mittag-Leffler)

Let $(X_\alpha, f_{\alpha\beta})$ be an inverse system of complete Hausdorff uniform spaces, indexed by a directed set $I$ which has a countable cofinal subset; suppose also that, for each $\alpha \in I$, $X_\alpha$ has a countable fundamental system of entourages (*). Finally, suppose that for each $\alpha \in I$ there is an index $\beta \geq \alpha$ satisfying the following condition:

(ML$_{\alpha\beta}$) For each $\gamma \geq \beta$, $f_{\alpha\gamma}(X_\gamma)$ is dense in $f_{\alpha\beta}(X_\beta)$.

Let $X = \varprojlim X_\alpha$ and let $f_\alpha$ be the canonical mapping $X \to X_\alpha$. Then for each $\alpha \in I$ and each $\beta \geq \alpha$ satisfying (ML$_{\alpha\beta}$), $f_\alpha(X)$ is dense in $f_{\alpha\beta}(X_\beta)$ (and consequently $X$ is non-empty if the $X_\alpha$ are all non-empty).

Let $(\lambda_n)$ be a sequence of indices cofinal in $I$. Start with an index $\alpha_0 \in I$ and define recursively an increasing sequence $(\alpha_n)$ such that $\alpha_n \geq \lambda_n$ and such that (ML$_{\alpha_n, \alpha_{n+1}}$) is true. Clearly the sequence

(*) This condition signifies that the Hausdorff uniform space $X_\alpha$ is metrizable; cf. Chapter IX, § 2, no. 4, Theorem 1.

(α_n) is cofinal in I. We shall write f_{mn} in place of f_{α_mα_n} for m ≤ n, and we shall put f_{n, n+1}(X_{α_{n+1}}) = Y_n. Then, if m ≤ n, f_{mn}(Y_n) is dense in Y_m; for by definition f_{m, n+1}(X_{α_{n+1}}) is dense in

f_{m, m+1}(X_{α_{m+1}}) = Y_m, and f_{m, n+1}(X_{α_{n+1}}) = f_{mn}[f_{n, n+1}(X_{α_{n+1}})] = f_{mn}(Y_n).

By induction on n and k we can define a fundamental system (V_{kn})_{k∈\mathbf{N}} of closed symmetric entourages of X_{α_n} for each n such that

(2) $\overset{2}{V}_{k+1, n} \subset V_{kn}$,

(3) $(f_{n, n+1} \times f_{n, n+1})(V_{k, n+1}) \subset V_{kn}$.

In effect, let (U_{kn})_{k∈\mathbf{N}} be a fundamental system of entourages of X_{α_n}. If we suppose that the V_{kn} have been defined for a given n and for all k ∈ \mathbf{N}, then since f_{n, n+1} is uniformly continuous we can define the entourage V_{k, n+1} by induction on k so that (3) is satisfied and

$\overset{2}{V}_{k+1, n+1} \subset V_{k, n+1} \cap U_{k+1, n+1}$

The assertion follows.

Now let x_0 ∈ Y_0. We shall show that for each integer k > 0 there is a point z ∈ X such that [x_0, f_{α_0}(z)] ∈ V_{k-1, 0}; this will prove the theorem. Since f_{n, n+1} (Y_{n+1}) is dense in Y_n, we can define by induction a sequence of points x_n ∈ Y_n such that

(4)

By reason of (3) it follows that if m ≤ n then

(5)

From this we conclude that for fixed m the sequence (f_{mn}(x_n))_{n≥m} is a Cauchy sequence in X_{α_m} and therefore converges to a point z_m; for by induction it follows from (5) that, for each pair of integers p ≥ m, q > 0, we have

(6)

and by virtue of (2) it is clear that the right-hand side of (6) is contained in V_{k+p-1, m}. Let q increase indefinitely; we infer in particular that, for m = p = 0, we have (x_0, z_0) ∈ V_{k-1, 0}, since V_{k-1, 0} is closed. On the other hand, from the relations z_m = \lim_{n→∞} f_{mn}(x_n) and from the continuity of f_{m, m+1}, we deduce that f_{m, m+1}(z_{m+1}) = z_m for each m ≥ 0. For each γ ∈ I there is at least one integer n such that α_n ≥ γ; putting $z_\gamma = f_{\gamma, \alpha_n}(z_n)$ we verify immediately that $z_\gamma$ does not depend on the value of $n$ such that $\alpha_n \geq \gamma$, and that the family $(z_\alpha)_{\alpha \in I}$ so defined is a point $z$ of $X = \varprojlim X_\alpha$. Since $f_{\alpha_0}(z) = z_0$, the proof is complete.