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This is a problem in Lee's Introduction to Smooth Manifolds.

Show that a disjoint union of uncountably many copies of $\mathbb{R}$ is not second-countable.

Let $S$ be the disjoint union of uncountably many copies of $\mathbb{R}$. It suffices to construct an open subset $U\subset S$ that cannot be written as the union of countably many open sets in $S$. Somehow, I reckon that one can obtain a contradiction by forming a bijection between the index set of the disjoint union (which is uncountable) and the claimed basis (which is countable).

I try reasoning as follows. Let $\mathcal{B}=\{B_1,B_2,\dots\}$ be a claimed countable basis. Let $A$ be the index set of $S$. Let $A_i\subset A$ be the set of indices $\alpha\in A$ such that there exists an element of the form $(x,\alpha)\in B_i$. Observe that if each $A_i$ was countable, then $A'=\bigcup_{i=1}^{\infty}{A_i}$ is countable as a countable union of countable sets. Since $A$ is uncountable, we must have that $A'$ is a proper subset of $A$. So there exists $\alpha_0\in A\setminus A'$, and now the set $$\{(x,\alpha_0)\colon x\in T\}$$ where $T$ is open in $\mathbb{R}$, is an open set in $S$ that cannot be written using the basis $\mathcal{B}$, a contradiction. Hence, there exists an integer $n$ such that $A_n$ is uncountable.

But I am not sure where to go from here.

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  • $\begingroup$ Getting one set isn't really a problem. $\endgroup$
    – user403337
    May 4 at 2:28
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The copies of $\mathbb R $ themselves form a collection of open sets that can't be written as the union of countably many open sets. For they form a disjoint collection of cardinality $\ge\mathfrak c $.

To pan it out, each copy of $\mathbb R $ would have to contain an element of the base. But that makes uncountably many (elements of the base).

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  • $\begingroup$ Each copy of $\mathbb{R}$ must contain a set in the basis? Why is that true? For instance, $$\{(x,\alpha_0)\colon0<x<1\}\cup\{(x,\alpha_1)\colon0<x<1\}$$ where $\alpha_0,\alpha_1\in A$ is open in $S$, but this set has "intersection" with two different copies of $\mathbb{R}$. Similarly, can't the sets in the basis "intersect" more than one copy of $\mathbb{R}$? $\endgroup$ May 4 at 4:34
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    $\begingroup$ Isn't each copy of $\mathbb R$ open? $\endgroup$
    – user403337
    May 4 at 4:36
  • $\begingroup$ Right, but I'm not understanding how we are using that. $\endgroup$ May 4 at 4:41
  • $\begingroup$ Doesn't every open set have to contain an element of the base? $\endgroup$
    – user403337
    May 4 at 4:42
  • $\begingroup$ Ohhh right that is true. Otherwise, there is an open set that cannot be written as the union of any of the basis sets. Thank you! $\endgroup$ May 4 at 4:43

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