If $X^n$ is Lindelöf, then so is $\widetilde{X}^n$. Let $G$ be a topological group $T_2$, and $X$ a closed subspace of $G$. We suppose that, for each $n\in\mathbb{N}$, $X^n$ is Lindelöf.
Consider $\widetilde{X}:=X\oplus\{e\}\oplus X^{-1}$ (we will assume that $x\neq e$ and $x\neq x^{-1}$ for each $x\in X$, thus these sets are disjoint).
In a book, the autor states (without a proof) that:
"For each $n\in\mathbb{N}$, $\widetilde{X}^n$ is Lindelöf because $\widetilde{X}^n$ is a finite union of closed copies of the spaces $X^k$, with $k\le n$."
So, I don't understand two things:
1.- Why $\widetilde{X}^n$ is a finite union of closed copies of the spaces $X^k$, with $k\le n$?
2.- Why this implies that $\widetilde{X}^n$ is Lindelöf?
Thanks.
 A: Given a space $Y$ and a positive integer $n$, let me write $nY$ for the discrete union of $n$ copies of $Y$. I’ll also write $\mathbf{1}$ for the one-point space.
First note that $X^{-1}$ is homeomorphic to $X$, so $\widetilde{X}$ is homeomorphic to $2X\oplus\mathbf{1}$. Let $Y=2X$. Then $\widetilde{X}^n$ is homeomorphic to $(Y\oplus\mathbf{1})^n$. If you write out the Cartesian product, you’ll find that
$$(Y\oplus\mathbf{1})^n=\bigoplus_{k=0}^n\binom{n}k\left(Y^k\times\mathbf{1}^{n-k}\right)\;,$$
which is clearly homeomorphic to $$\bigoplus_{k=0}^n\binom{n}kY^k\;.$$ And
$$Y^k=(X\oplus X)^k=\bigoplus_{j=0}^k\binom{k}j\left(X^j\times X^{k-j}\right)=\bigoplus_{j=0}^k\binom{k}jX^k=2^kX^k\;,$$
so $\widetilde{X}^n$ is homeomorphic to $$\bigoplus_{k=0}^n\binom{n}k2^kX^k\;,\tag{1}$$
the discrete union of a finite number of copies of $X^k$ for $0\le k\le n$. (Note that $X^0$ is simply the space $\mathbf{1}$.)
By hypothesis each of the spaces $X^k$ with $k\le n$ is Lindelöf. Thus, if $\mathscr{U}$ is an open cover of the space $(1)$, then each of the subspaces $X^k$ can be covered by countably many members of $\mathscr{U}$. There are only finitely many of these subspaces, and the union of finitely many countable sets is still countable, so only countably many members of $\mathscr{U}$ are required to cover the space $(1)$. This space, and hence the homeomorphic space $\widetilde{X}^n$, is therefore Lindelöf.
We don’t actually need to know that the copies of $X^k$ are closed in $(1)$ (though of course they are both closed and open), and it would have been sufficient to know that there are countably many of them (rather than just finitely many): the argument that I just gave to show that the space $(1)$ is Lindelöf works whether or not the subspaces $X^k$ are closed, and it works even if there are countably infinitely many of them instead of finitely many.
