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[submodule "LatexPackagesBuild"]
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[submodule "LatexPackagesBuild"]
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path = LatexPackagesBuild
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path = LatexPackagesBuild
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url = https://gitlab.com/latexci/packages/LatexPackagesBuild.git
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url = https://git.abstractnonsen.se/latex/latex-packages-build
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@ -160,14 +160,14 @@ Let $I$ be a linear order
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we have that $F_\gamma = \bigcap_{\alpha < \gamma} F_\alpha$,
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we have that $F_\gamma = \bigcap_{\alpha < \gamma} F_\alpha$,
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since $(X_\gamma,\Z)$ is the inverse limit of
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since $(X_\gamma,\Z)$ is the inverse limit of
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$\{(X_{\alpha}, \Z):\alpha < \gamma\}$.
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$\{(X_{\alpha}, \Z):\alpha < \gamma\}$.
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\item For all $\alpha < \beta$, $F_{\alpha+1} \subsetneq F_\alpha$,
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\item For all $\alpha$ it is $F_{\alpha+1} \subsetneq F_\alpha$,
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because $\pi^{\alpha+1}_\alpha \colon (X_{\alpha+1},\Z) \to (X_\alpha,\Z)$
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because $\pi^{\alpha+1}_\alpha \colon (X_{\alpha+1},\Z) \to (X_\alpha,\Z)$
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is not a bijection
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is not a bijection
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and all the fibers are isomorphic.
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and all the fibers are isomorphic.
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\end{itemize}
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\end{itemize}
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So $(F_\alpha)_{\alpha \le \beta}$ is a strictly
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So $(F_\alpha)_{\alpha \le \beta}$ is a strictly
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increasing chain of closed subsets.
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decreasing chain of closed subsets.
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But $X$ is second countable,
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But $X$ is second countable,
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so $\beta$ is countable:
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so $\beta$ is countable:
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Let $\{U_n\} = \cB$ be a countable basis
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Let $\{U_n\} = \cB$ be a countable basis
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