show that A is a locally compact in the product topology

Let $X=[-1,1]^N$ have the product topology, where each interval has the usual topology. Let $A$ be the subset of $X$ consisting of all sequences $(a_1, a_2 , ...)$ for which $a_i=2a_{i+1}^2 -1$, $i \in N$ and $a_i \in [-1,1]$. Prove that A is locally compact.

proof:-

Since each copy of $[-1,1]$ is a compact Hausdorff space, then the product $[-1,1]^N$ is a compact Hausdorff space.

Now, we want to show that $A$ is a closed subset. Let $(x_n)$ is a sequence in $A$, and $(x_n) \to x=(a_1,a_2,a_3,...)\in X$, we can write $$x_n=(a_{n,1},a_{n,2},a_{n,3},...)\to x=(a_1,a_2,a_3,...)$$ For fixed $i$, by using the continuity of the projection map $\pi_i$, $$a_{n,i}=\pi_i(x_{n})= \pi_i (a_{n,1},a_{n,2},a_{n,3},...) \to \pi_i(x)=a_i$$ Since $i$ is arbitrary, then $\lim_{n\to \infty }a_{n,i}=a_i$ for all $i$

But $a_{n,i}=2a_{n,i+1}^2-1$, so $$\lim_{n\to \infty }2a_{n,i+1}^2-1=a_i$$ Thus $$2a_{i+1}^2-1 = a_i$$ Therefore, $x \in A$.

Hence, $A$ is closed in a compact Hausdorff space, so $A$ is a compact Hausdorff.

Therefore $A$ is a locally compact.

Notes:-

Definition: A space $X$ is locally compact iff each point in $X$ has a nhood base consisting of compact sets.

Theorem1: A Hausdorff space $X$ is locally compact iff each point in $X$
has a compact nhood.

Theorem2: In a locally compact Hausdorff space, the intersection of an open set with a closed set is locally compact. Conversely, a locally compact subset of a Hausdorff space is the intersection of an open set and a closed set.

Theorem3: If f is a continuous, open map of $X$ onto $Y$ and $X$ is locally compact, then so is $Y$.

Theorem4: Suppose $X_a$ is nonempty for each $α \in A$. Then $\prod X_a$ is locally compact iff

a) each $X_a$ is locally compact,

b) all but finitely many $X_a$ are compact.

I'm confused, where is the direction that leads me to the result.

Any help would be very appreciated.

This space is just the inverse limit of a sequence of copies of $[-1,1]$ with bonding maps $f(x)=2x^2-1$. This is a compact Hausdorff space, as a closed subspace of $[-1,1]^{\mathbb{N}}$, and so also locally compact.
• Define $A_k=\{(x_n)_n\mid x_k= f(x_{k+1})\}$. Show that all $A_k$ are closed, $k=1,2,3,\ldots$. You set is $\bigcap_k A_k$. So closed in a compact Hausdorff space. – Henno Brandsma Nov 25 '17 at 21:51
• I try to show that $A_k$ is closed, but I can not !! – Hamada Al Nov 29 '17 at 4:21