Example of a Hausdorff topology that is homeomorphic to $\mathbb R^n$ but isn't second countable? The definition of a topological manifold $M$ I have is:

*

*$M$ is Hausdorff.


*Each point of $M$ has a neighborhood
that is homeomorphic to an open subset of $\mathbb{R}^n$.


*$M$ is second countable.
That last one confuses me a little since it seems to me the second condition should imply the third.
$\mathbb{R}^n$ is second countable, thus an open subset $S$ of it is also second countable through the base:
$B_S = \{U \cap S : U \in B; S \subseteq \mathbb{R}^n\}$ where $B$ is the base of $\mathbb{R}^n$,
In English, grab the base for $\mathbb{R}^n$, for each set in it, intersect it with the open space, append all the results to a new set. By construction this new set is both countable and any open set in $S$ is a union of sets in $B_S$.
And now use the homeomorphism $\phi$ to map the base sets onto $M$, Since we used a homeomorphism the resulting set $\phi(B_S)$ should be a countable base for $M$.
Where is my logic failing? What is an example of a topology that ahs the first 2 properties without the third?
 A: I see three issues with your argument.

*

*You seem to show that every point has a neighborhood with a countable base.  That is not the definition of "second countable."  [This part is true, though.]

*You "complete" the argument by doing this with every chart $\phi$.  But why must all these charts be countable in number?  Typically they aren't.  So when you do this for "all" charts, there is no guarantee that the end result is still countable.

*You say that you can use the charts to map these bases "onto" $M$, but charts won't be onto generally.

In short, nothing about this gives a countable basis for the topology of $M$.
A: The second condition does not say that the whole space is homeomorphic to $\mathbb R^n,$ but rather that each point has some open neighborhood that is homeomorphic to an open subset of $\mathbb R^n.$
Your subject line, as it appears now, says:

Example of a hausdorff topology that is homeomorphic to $\mathbb R^n$ but isn't second countable?

Every Hausdorff space that is homeomorphic to $\mathbb R^n$ satisfies the second axiom of countability. But that does not mean that every Hausdorff space in which every point has an open neighborhood homeomorphic to some open subset of $\mathbb R^n$ satisfies the second axiom of countability.
One example is the "long line," unless maybe I should say "long ray."
Recall that a well-ordered set is one that is linearly ordered in such a way that every non-empty subset of it has a smallest member. For present purposes we can take the term "ordinal number" mean the order type, i.e. the isomorphism class, of some well-ordered set.
Now consider the set of all countable ordinals. The set of all countable ordinals is uncountable. Its cardinality is the one to which Georg Cantor gave the name $\aleph_1.$ Now between each ordinal and it successor, glue a copy of the open unit interval $(0,1).$ There you have a Hausdorff space, the long line or long ray, in which every point except $0$ has a neighborhood homeomorphic to the real line. And if you just discard $0$ from the space, that's your example: it does not satisfy the second axiom of countability.
