First Countability, Continuous f:[a,b]->[0,1]

joypav

Active member
Lemma: (Professor said we would be able to use this for the problem? But we have to prove it to use it.)
If X is a regular Hausdorff space and X is first countable at the point P. Then there is a local basis $\left\{B_i\right\}^{\infty}_{i=1}$ at P so that for each $n \in \Bbb{N}$ we have:
$\overline{B_{n+1}} \subset B_n$.

Proof of Lemma:
X first countable at P $\implies \exists$ a countable local basis $\left\{A_i\right\}^{\infty}_{i=1}$ at P.
Let $B_n(P) = A_1 \cap A_2 \cap A_3 \cap ... \cap A_n$.

Then $\left\{B_i(P)\right\}^{\infty}_{i=1}$ is a collection of open sets such that
$B_{i+1}(P) \subset B_i(P)$, for all i.

Need to Show: $\left\{B_i(P)\right\}^{\infty}_{i=1}$ is a local basis at P.

Consider an open set B containing P.
$\left\{A_i\right\}^{\infty}_{i=1}$ is a local basis at P and B is an open set containing P $\implies$ (by first countability)
$\exists i_0 \in \Bbb{N}$ such that $A_{i_0} \subseteq B$.

By definition of the $B_n(p)$'s,
$A_1 \cap A_2 \cap ... \cap A_{i_0} = B_{i_0}(p) \subseteq A_{i_0} \subseteq B$

So for any open set B containing P, $\exists i_0 \in \Bbb{N}$ so that $B_{i_0}(P) \subseteq B$.

$\implies \left\{B_i(P)\right\}^{\infty}_{i=1}$ is a local basis at P.

Need to Show: for each $n \in \Bbb{N}, \overline{B_{n+1}} \subset B_n$

X regular $\implies$ if $P \in B_i(P) \implies \exists U$ open in X such that $P \in U \subset \overline{U} \subset B_i(P)$

However, $\left\{B_i(P)\right\}^{\infty}_{i=1}$ is a local basis at P $\implies$
$\exists B_m(P)$ so that $P \in B_m(P) \subset U$
$\implies P \in B_m(P) \subset \overline{B_m(P)} \subset U \subset B_i(P)$
Then, $\overline{B_m(P)} \subset B_i(P)$.

I'm sort of stuck here... is this the right idea for the proof or not so much?

Problem 1.
Suppose that X is a linearly ordered space that has first and last points a and b respectively. Suppose further that X is first countable at both endpoints. Show that there is a continuous function $f : X \rightarrow [0, 1]$ so that $f^{−1}(0) = {a}$ and $f^{−1}(1) = {b}$.

Proof:

I chose the function
$f(x) = \frac{x - a}{b - a}$

I think this should work fine. Do I just need to prove continuity now?

joypav

Active member
I guess my question for the problem is, why do I need first countability at each endpoint?

Opalg

MHB Oldtimer
Staff member
Problem 1.
Suppose that X is a linearly ordered space that has first and last points a and b respectively. Suppose further that X is first countable at both endpoints. Show that there is a continuous function $f : X \rightarrow [0, 1]$ so that $f^{−1}(0) = {a}$ and $f^{−1}(1) = {b}$.

Proof:

I chose the function
$f(x) = \frac{x - a}{b - a}$

I think this should work fine. Do I just need to prove continuity now?
The problem here is that you seem to be assuming that $X$ is an interval in the real line. But you are only told that $X$ is a linearly ordered space. There is no mention of any additive structure on $X$. In that case, how are you going to define "$x-a$" and "$b-a$"?