Is the product space of a non-second-countable space and a second-countable space non-second-countable?

Question

I am asked if $\mathbb{R} \times [0, 1]$ would be a manifold if $\mathbb{R}$ is defined with lower limit topology and $[0, 1]$ is defined with subspace topology for standard topology. Although $[0, 1]$ is second-countable, I think that the product space won't be second-countable since $\mathbb{R}$ is not second-countable. Therefore, the product can't be a manifold. If I am thinking in the right way, how can I show this?

Answer 1

Suppose $f:X\to Y$ is continuous, open and surjective. If $X$ is second countable, then so is $Y$. Indeed, let $\mathcal{B}$ be a countable basis for $X$. Then $$ \{f(B):B\in \mathcal{B}\} $$ is a basis for the topology of $Y$. To see this, let $U\subset Y$ be open and pick $y\in U$. Since $f$ is surjective, $y=f(x)$ for some $x\in X$. Obviously, $x\in f^{-1}(U)$, which is open, so $x\in B\subset f^{-1}(U)$ for some $B\in\mathcal{B}$. But then $y=f(x)\in f(B)\subset U$, and we're done.

Remark: the fact that $f$ is an open function was used to conclude that each $f(B)$ is open in $Y$.

---

Now apply this to the projection $\mathbb{R}\times [0,1]\to \mathbb{R}$.