Why is this space not second-countable?

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.

• Getting one set isn't really a problem.
– user403337
May 4 at 2:28

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).

• 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 $\mathbb{R}$? May 4 at 4:34
• Isn't each copy of $\mathbb R$ open?
– user403337
May 4 at 4:36
• Right, but I'm not understanding how we are using that. May 4 at 4:41
• Doesn't every open set have to contain an element of the base?
– user403337
May 4 at 4:42
• 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! May 4 at 4:43