Prove $ℝ^I$ is not Lindelöf Munkres's topology book has a large review exercise, which includes testing topological axioms on a variety of spaces. One of those spaces is $ℝ^I$ where $I=[0,1]$ is the unit interval, and one of the properties is the Lindelöf condition. I can deduce that $ℝ^I$ isn't Lindelöf seeing as it is regular, as the two would imply paracompactness, implying normality, which it fails to satisfy. But how can one prove this result without referring to paracompactness?
I can tell that $ℝ^I$ being Lindelöf would imply the closed $ℤ^I⊆ℝ^I$ is as well, but I'm not sure even how to work that case, if it even isn't Lindelöf.
 A: Your argument doesn’t need paracompactness: just show directly that every regular Lindelöf space is normal. This is fairly straightforward; see, for instance, Theorem $16.8$ in Willard’s General Topology.
As you say, if $\Bbb R^I$ were Lindelöf, $\Bbb Z^I$ would also be Lindelöf, and I’ll show directly that it’s not. Let $\mathscr{P}=\{P\subseteq I:|P|=2\}$. For each $P=\{x,y\}\in\mathscr{P}$ let $U(P)=\{p\in\Bbb Z^I:p(x)=p(y)\}$, and let $\mathscr{U}=\{U(P):P\in\mathscr{P}\}$; since $|I|>|\Bbb Z|$, every point of $\Bbb Z^I$ is in some $U(P)$, and $\mathscr{U}$ is an open cover of $\Bbb Z^I$.
Let $\mathscr{V}=\{U(P_n):n\in\Bbb N\}$ be a countable subset of $\mathscr{U}$. Let $A=\bigcup_{n\in\Bbb N}P_n$. Then $A$ is countable, so there is an injection $f:A\to\Bbb Z$. Let
$$p:I\to\Bbb Z:x\mapsto\begin{cases}
f(x),&\text{if }x\in A\\
0,&\text{otherwise}\;;
\end{cases}$$
clearly $p\in\Bbb Z^I\setminus\bigcup\mathscr{V}$, so $\mathscr{V}$ fails to cover $\Bbb Z^I$. Thus $\mathscr{U}$ has no countable subcover, and $\Bbb Z^I$ is not Lindelöf.
