# Can the long line be embedded in euclidean space?

In the definition of a manifold $M$, we have the following conditions:

1. For some fixed $n$, $M$ is locally homeomorphic to $\mathbb{R}^d$.
2. $M$ is connected, second countable, and Hausdorff.

Now, with this definition, it is a well-known theorem that $M$ can be embedded in $\mathbb{R}^{2n+1}$. I suspect that if the condition of second countability is relaxed, then this theorem is no longer true, and by attempt at a counterexample is to define $L=\omega_1\times[0,1)$ with the order topology, where $\omega_1$ is the first uncountable ordinal. $L$ can be seen to be a non-second-countable manifold of dimension one, however, I'm struggling to prove that $L$ cannot be embedded in $\mathbb{R}^3$. I have a vague feeling that by using uncountably many points in general position and well-ordering, an embedding can in fact be constructed, though I may be wrong about this.

If $L$ is not a counter-example to the hypothesis that non-second-countable manifolds can still be embedded in Euclidean space, then what is?

No, because of the Poincaré-Volterra theorem. Here's the statement you can find in Bourbaki's General Topology (I.11.7, corollary 3).

Theorem. Let $Y$ be a locally compact, locally connected space whose topology has a countable base. Let $X$ be a connected Hausdorff space, and let $p : X \to Y$ be a continuous mapping which has the following property: for each $x \in X$ there is an open neighbourhood $U$ of $x$ in $X$ such that the restriction of $p$ to $U$ is a homeomorphism of $U$ onto an open subspace of $Y$. Then $X$ is locally compact and locally connected, and the topology of $X$ has a countable base.

A classical use of this theorem is to show that Riemann surfaces are automatically secound countable, even if you don't make it an explicit hypothesis (the general theory shows that there is a nonconstant holomorphic map $S \to \mathbb P^1(\mathbb C)$ and you apply the Poincaré-Volterra theorem).

So, for a manifold, being secound-countable is equivalent to being embeddable in some Euclidean space. These are two of the 119 (!) equivalent conditions you can find on theorem 2.1 of Gauld's Non-metrisable manifolds.

• This is an incredibly small nitpick of a very good answer -- at least in the 2014 edition of Gauld's book, Theorem 2.1 gives equivalent conditions for a manifold to be metrizable, which is strictly weaker than being second-countable. For instance, an uncountable discrete space is metrizable (discrete metric) but clearly not second countable and thus cannot be embedded into any Euclidean space. – Chill2Macht Mar 2 '17 at 11:08

Recall that every subspace of a second-countable space is also second-countable. (If $\mathcal B$ is countable base for $X$ then $\{ U \cap Y : U \in \mathcal B \}$ is a countable base for $Y \subseteq X$.) Therefore no non-second-countable space (manifold or otherwise) can embed in any Euclidean space.

• Ah, so this provides yet another example of a continuous injection that is not an embedding. I see the flaw in my original chain of thought. – Monstrous Moonshine Aug 16 '16 at 22:46