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.


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.


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.

  • $\begingroup$ Should I use Theorem 3 or what?? $\endgroup$ – Hamada Al Nov 25 '17 at 21:40
  • $\begingroup$ Compact implies locally compact is trivial from the definition. @HamadaAl it also follows from theorem 2. $\endgroup$ – Henno Brandsma Nov 25 '17 at 21:41
  • $\begingroup$ Oh yes, you mean compact Hausdorff space gives locally compact. But could you clarify to me why A is compact Hausdorff ? $\endgroup$ – Hamada Al Nov 25 '17 at 21:45
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Nov 25 '17 at 21:51
  • $\begingroup$ I try to show that $A_k$ is closed, but I can not !! $\endgroup$ – Hamada Al Nov 29 '17 at 4:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.