1

Given a set $X$ and a topological space $Y$ the set of all (not necesarilly continuous) maps from $X$ to $Y$ is denoted by $Y^X$.

As a brief introduction to the problem, I first want to specify what I mean by normal space as I'm not sure if theres different deffinitions.

A topological space $X$ is $T_1$ if for every pair of different points $x,y\in X$ there exist two open sets $U,V$ sutch that $x\in U$, $x\notin V$ and $y\in V$, $y\notin V$.

A topological space $X$ is normal if it is $T_1$ and if for every two disjoint closed sets $C_1,C_2 \subset X$ there exist two disjoint open sets $U_1,U_2$ sutch that $C_1\subset U_1$ and $C_2 \subset U_2$.

Now, over the set of all maps from $X$ to $Y$ the point-open topology is the one defined by the subbase $$\mathcal S = \{U^x:x\in X \textrm{ and }U\textrm{ is open in }Y\}$$ where $A^B = \{f\in X^Y : f(B)\subset A\}$ is the notation used in this post where $A\subset Y$ and $B\subset X$.

With this in mind, in the book "Topology and Maps" by T.Husain, in chapter VIII there is an exercise that says:

If $E$ is a set and $F$ is a normal topological sapce, then $F^E$ is not necessarily normal.

I've been thinking about such a space but I can't find an example.

Mikel Solaguren
  • 495
  • Which spaces have you tried? – CyclotomicField May 07 '25 at 19:35
  • @CyclotomicField I first tried with $\mathbb R ^\mathbb{R}$ taking the closed sets of the form ${0}^\mathbb{R}$ and ${1}^\mathbb{R}$ but there exist $(-\tfrac{1}{2},\tfrac{1}{2})^{0}$ and $(\tfrac{1}{2},\tfrac{2}{3})^{0}$ which are disjoint and open sets containing each. Then I tried with some "weirder" spaces like ${1,2}^\mathbb{R}$ but I realized that when $Y$ has the discrete topology the function space is always normal because closed sets are also open and thus given two disjoint closed sets $C_1^X$, $C_2^X$ those sets are also open and disjoint. – Mikel Solaguren May 07 '25 at 20:23
  • In this answer Brian Scott proves that $\mathbb{N}^{\omega_1}$ is not normal.

    As a corollary, if $F^E$ is normal and $E$ is uncountable, then $F$ is limit point compact.

     – Jakobian May 08 '25 at 14:09

1 Answers1

2

I think the use of the phrase "compact-open topology" is a bit misleading here. Usually one works with topological spaces $X, Y$ and endows the set $C(X,Y)$ of continuous maps $X \to Y$ with the compact-open topology.

You consider a set $X$ and a topological space $Y$ and endow the set $F(X,Y)$ of all functions $X \to Y$ with the topology generated by the subbase $\mathcal S$. Actually this is the topology of pointwise convergence on $F(X,Y)$.

However, if you want, you can endow the set $X$ with the discrete topology . Then $F(X,Y) = C(X,Y)$ and the topology of pointwise convergence on $F(X,Y)$ agrees with the compact-open topology on $C(X,Y)$.

Anyway, it is well-known that we can identify $Y^X$ with the product $\prod_{x \in X} Y_x$, where $Y_x = Y$ for all $x \in X$.

Now let $F$ be the Sorgenfrey line. Its underlying set is $\mathbb R$ and as a base one takes the collection of half-open intervals $[a, b)$. It is well-known that $F$ is normal. However, the product $F \times F \approx F^{\{1,2\}}$ is not normal. See Is there a simple method to prove that the square of the Sorgenfrey line is not normal?

Paul Frost
  • 87,968