Existence of limit point & compactness

Question

Let $\tau =\{\varnothing\}\cup\{\mathbb{R}-X\,|\, X\text{ is countable}\} $ be a topology on $\mathbb{R}$. Consider the product $Y=\mathbb{R}\times\{0,1\}$ with the product topology, where $\{0,1\}$ is discrete.

1. Show that any proper infinite subspace of $Y$ has a limit point.
2. Is $Y$ compact?

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I honestly cannot wrap my head around this. Any tips are welcome!

Answer 1

1. $A=\Bbb N\times \{0\}$ has no limit points. In fact, let $x$ be a limit point. Since $A$ is closed, $x=(m,0)\in A$. On the other hand, if $x\in A$, then $U=((\Bbb R\setminus \Bbb N)\cup\{m\})\times \{0,1\}$ is a neighbourhood of $x$ such that $U\cap A=\{(m,0)\}$.

2. $(\Bbb R,\tau)\times \{0,1\}$ is compact if and only if either all factors are compact or one factor is empty. Both factors are non-empty and $(\Bbb R,\tau)$ is not compact because the family of closed sets $\mathcal F=\{[n,\infty)\cap \Bbb N\,:\, n\in\Bbb N\}$ satisfies $\bigcap \mathcal F=\emptyset$, while no finite subfamily has empty intersection. Therefore $Y$ is not compact.

Answer 2

$2:$ To see that $Y$ is not compact, we need to create a cover that has no finite subcover. Let $U_1$ be an open set in $Y$ such that $U_1 = \mathbb{R}-X \times \{0,1\}$ where $X$ is a countably infinite set. Ennumerate the elements of $X$, $\{x_i\}_{i=1}^{\infty}.$

Then $(\mathbb{R}-X)^c = X$ is a countable infinite set. Choose $U_2$ to be $U_1 \cup (\{x_1\} \times \{0,1\})$. So $U_2$ is just $U_1$ and we added in the first element of the ennumaration of $X$. Continuing this way we will build is an infinite sequence of nested intervals (which means $U_k \subset U_{k+1}$ for all $k \in \mathbb{N}$) with no finite subcover. Thus $Y$ is not compact.