Hierarchy of Conditions on Locally Euclidean Spaces (8.50.Supp 2-6)

James Munkres Topology, chapter 8.50, supplementary exercises 2-6:

Let $X$ be locally $m$-euclidean.

2. Consider the following conditions on $X$:
(i) $X$ is compact Hausdorff (ii) $X$ is an $m$-manifold (iii) $X$ is metrizable (iv) $X$ is normal (v) $X$ is Hausdorff

Show (i) $⇒$ (ii) $⇒$ (iii) $⇒$ (iv) $⇒$ (v).

3. Show $ℝ$ is locally $1$-euclidean and satisfies (ii) but not (i).

4. Show that $ℝ×ℝ$ in the dictionary order topology is locally $1$-euclidean and satisfies (iii) but not (ii).

5. Show that the long line is locally $1$-euclidean and satisfies (iv) but not (iii).

Proof: 2. [(i) $⇒$ (ii)] Since $X$ is locally metrizable, Hausdorff compactness of $X$ implies by the Smirnov metrization theorem that $X$ is metrizable. Since compact metric spaces are second countable, it follows $X$ is an $m$-manifold. [(ii) $⇒$ (iii)] This follows by the Urysohn metrization theorem. [(iii) $⇒$ (iv)] Metric spaces are necessarily normal. [(iv) $⇒$ (v)] Normal spaces are necessarily Hausdorff.

3. $ℝ$ is clearly Hausdorff locally $1$-euclidean with a countable basis, but is not compact.

4. It is clear $ℝ×ℝ$ in the dictionary order topology is locally $1$-euclidean and does not have a countable basis, seeing as $r×(0,1)$ for $r∈ℝ$ is an uncountable collection of disjoint nonempty open sets in $ℝ×ℝ$. However, by paracompactness of $ℝ$, it is seen that $ℝ×ℝ$ is paracompact, so that by the Smirnov metrization theorem the space is metrizable.

5. Every order topology is normal, and by the results of the exercises of chapter 26 we know that the long line is locally $1$-euclidean. However, the long line is limit point compact but not compact, so that it cannot be metrizable.$~\square$