Showing that $X$ and the inverse limit of the sequence $\mathcal{L}$ are homeomorphic.

Question

Having known how to solve the following problem:

Let $X$ be totally disconnected, compact, metric and let $\mathcal{L}$ be the inverse limit of the sequence. (I mentioned in my previous question).

$(a)$ Write out what a typical element of $\mathcal{L}$ is in terms of subsets of $X$ and the inclusions of sets in other sets.

$(b)$ Find a way to uniquely determine a point $x_{\sigma} \in X,$ given an element $\sigma \in \mathcal{L}.$ This defines a function $f:\mathcal{L} \rightarrow X.$

$(c)$Find a way to uniquely determine a point $\sigma \in \mathcal{L}$ given a point $x_{\sigma} \in X.$ This defines a function $g: X \rightarrow \mathcal{L}.$

$(d)$ Show that $f$ and $g$ are mutual inverses. That is, show that $f \circ g = id_{X}$ and $g \circ f = id_{\mathcal{L}}.$

I want to solve the following problem:

Continuing with the notation and hypotheses of the previous problem:

(a) Show that $f$ is continuous iff $f^{-1}(U)$ is open in $\mathcal{L}$ for every $U \in \mathcal{A} = \bigcup_{n \geq \mathcal{A}_{n}}.$

(b) Show that $f$ is continuous.

(c) Deduce that $f$ is a homomorphism.

My questions are:

1- What is the difference between the proof of $(a)$ and $(b)$ in the problem that I want to solve?

2- How can I show that $g$ is continuous?

Could anyone help me answering those questions, please?

Could anyone help me in doing so, please?

Here is the question:

Henno Brandsma · Accepted Answer · 2020-04-13T09:23:43.673

0

In this anwer I showed that $\mathcal{A}$ is a base $X$.

This shows $(a)$, as continuity of a function only needs that the inverse elements from some base (or subbase) are open in the domain.

Now $f^{-1}[A]$, where $A \in \mathcal{A}_n$ for some $n$, is by definition of $A$, just equal to $p_n^{-1}[\{A\}]$, where the $p_n$ are the restricted projections from your linked question, onto the discrete finite spaces $\mathcal{A}_n$. So openness of $f^{-1}[A]$ follows, and so, by $(a)$ continuity.

$(c)$ is just applying the same fact as I did before, that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. QED

edited Apr 13 '20 at 09:23

answered Apr 12 '20 at 06:15

Henno Brandsma

242,131

Note that $g: X \to \mathcal{L}$ can be defined by $g(x)=(A_n)n$, where $A_n \in \mathcal{A}_n$ is the unique member that contains $x$, for each $n$. It's clear that $g(x) \in A\infty$ and that $f(g(x))=x$ as $x$ must be the unique intersection point, and this shows surjectivity of $f$ too. – Henno Brandsma Apr 13 '20 at 10:50
So, what is the difference between what is required in $(a)$ and what is required in $(b)$? Is your second paragraph from below a proof for $(b)$? – Apr 13 '20 at 11:19
for $(c)$ why $f$ is bijection here? Is it the identity function? – Apr 13 '20 at 11:20
@Happy yes and $g$ is its inverse. – Henno Brandsma Apr 13 '20 at 11:48
why it is the identity? – Apr 13 '20 at 11:53
$g \circ f$ is the identity on $\mathcal{L}$ and $f(g(x))$ is the identity on $X$. – Henno Brandsma Apr 13 '20 at 13:30
What about the answer to my first comment? – Apr 13 '20 at 14:10
Let us continue this discussion in chat. – Henno Brandsma Apr 13 '20 at 14:11

Showing that $X$ and the inverse limit of the sequence $\mathcal{L}$ are homeomorphic.

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