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Motivation: The line with two origins can be constructed via taking two copies of ${\mathbb R}$ and identifying $x$ on the first copy with $x$ on the second copy for each $x\neq 0$. We note that because this space is separable and first countable, it is also second countable. What if, instead of two origins, we have uncountably many? I read a claim that this space was not second countable, but I can't see why.

Problem: Define the line with uncountably many origins in the following way: take a disjoint union of uncountably many copies of ${\mathbb R}$; we may index them as ${\mathbb R}\times \{\alpha\}$ for $\alpha\in A$, an uncountable indexing set. We then define the equivalence class $\sim$ by $(x,\alpha)\sim (x,\beta)$ for $\alpha,\beta\in A$ and for all $x\neq 0$. Is the resulting quotient space second countable?

What I've Done: I know that the somewhat related problem of showing the disjoint union of an uncountable number of ${\mathbb R}$'s are not second countable relies on the fact that (after making it a metric space) one cannot find a countable dense subset --- but in this problem, I think, each open set covering $0\times \{\alpha\}$ will cover all of the other $0\times \{\beta\}$'s as well. In particular, I don't see why just taking all rational-centered balls with rational radii.

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This does not reduce to the problem of finding a countable dense subset, which is a weaker property known as "seperability". In the line with uncountably many origins, I can take an open set $U_{\alpha}$ that contains only the origin from $\mathbb{R} \times \{\alpha\}$, say the projection of the set $(-1,1) \subset \mathbb{R} \times \{\alpha\}$ to the quotient. If I have a basis for my topology, then I need an open set in my basis contained inside $U_{\alpha}$ that covers its origin, and so that basis element cannot have any other origins. Thus each of the $U_{\alpha}$ requires a basis element that cannot be used again for any other $U_{\alpha}$, necessitating uncountably many elements in my basis.

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    $\begingroup$ Ah, for some reason I was thinking that any open set containing one origin needed to contain them all; this isn't the case as you point out. This clears up the problem above -- thank you! $\endgroup$ – user2959 Feb 19 '13 at 4:54

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