real numbers have the topology with basis, $ [a,b)$ : Determining continuity on general polynomials

Question

Recently, I got the following problems from Kahn, [Topology, An Introduction to the point-set and algebraic areas]

Let the real numbers have the topology with basis, the sets $[a,b) = \{x \mid x\in \mathbb{R} , a\leq x <b\}$.

Is $f(x) = -x$ continuous?

The solution in the textbook says,

$f^{-1} ( [-1,+1)) = (-1,+1]$ and this cannot be open. If a basis set contained in $1$, $1 \in [a,b)$, it would contain some points bigger than $1$.

After reading this sentence, because of reversing $x\mapsto -x$, the above statement holds. Simply I can easily see any form of $cx$ with negative $c$ is not continuous. And also shifting gives the same answer, i.e., $x\mapsto cx+d$ with negative $c$ is not continuous.

Can it be generalized into a higher degree of polynomials? I mean $f(x) = cx^3 + dx^2 + ex + f$ or more higher degrees of polynomials.

Answer 1

$(1).$ The usual topological def'n of continuity is that $f:A\to B$ is continuous iff $f^{-1}C$ is open in $A$ whenever $C$ is open in $B.$ There are many useful equivalents to this. One is: $f:A\to B$ is continuous if $f(a)\in Cl_B (f(D))$ whenever $D\subset A$ and $a\in Cl_A(D).$

The name of the space in your Q is the Sorgenfrey Line (sometimes called the lower-limit topology on the set $\Bbb R$).

$(2).$ Let $f:\Bbb R\to \Bbb R$ be Real-continuous, that is, continuous with respect to the $usual$ (standard) topology on $\Bbb R.$ Suppose $a<b$ and $f(a)>f(b).$ Then let $a'=\sup \, ([a,b]\cap f^{-1}\{f(a)\})$ and $b'=\inf \,([a,b]\cap f^{-1}\{f(b)\}).$ We have

(i). $a\le a'<b'\le b.$

(ii). $f(a')=f(a)$ and $f(b')=f(b).$

(iii). $f((a',b'])=[f(b'),f(a')).$

Now let $D=(a',b'].$ Then $a'$ is in the Sorgenfrey-closure of $D.$ But $[f(a'),f(a')+1)$ is Sorgenfrey-open and disjoint from $f(D)=[f(b'),f(a')),$ so $f(a')$ is not in the Sorgenfrey-closure of $f(D).$ So $f$ is not Sorgenfrey-continuous.

So for a Real-continuous $f:\Bbb R\to \Bbb R$ to be Sorgenfrey-continuous, it is $necessary$ that $f$ is increasing. I.e., $a<b\implies f(a)\le f(b).$ I will leave it to you to show this is also $sufficient$ for a Real-continuous $f:\Bbb R\to \Bbb R$ to be Sorgenfrey-continuous

In particular, if $f$ is differentiable then $f$ is Sorgenfrey-continuous iff $f'(x)\ge 0$ for all $x.$

BTW. If $f'$ is a polynomial on $\Bbb R$ then $f'(x)\ge 0$ for all $x$ iff $f'$ is the sum of finitely many squares of polynomials.