Borel set in an uncountably product space with product topology

Question

Let $I=[0,1]$ be with its usual topology and $I^I$ be with product topology.

Let $1_A(x)$ be the indicator function of set $A$, i.e., $1_A(x)=1$ for $x\in A$ and $1_A(x)=0$ for $x\notin A$.

Let $f(x):=1_{\{x\}}$, so that $f$ defines a function from $I$ into $I^I$. Show that the range of $f$ is a Borel set in $I^I$.

My efforts:

For any $x\in I$, $U_x:=\{g\in I^I:g(x)\neq0\}$ is open. Thus $U_x^C:=\{g\in I^I:g(x)=0\}$ is closed and Borel. Similarly, $V_x^C:=\{g\in I^I:g(x)=1\}$ is closed and Borel.

Since arbitrary intersection of closed sets is closed, $V_y^C\cap\bigcap_{x\in I,x\neq y}U_x^C=\{1_{\{y\}}\}$ is closed and Borel for any $y\in I$.

Then I don't know how to continue.

Answer 1

Let the image of $f$ be $A$. We will show that $A$ is closed and hence Borel.

To show $A$ is closed, we will show $A$ equals its closure, i.e., $A=\bar{A}$.

To show $A=\bar{A}$, we will show: if $y\in\bar{A}$, then $y\in A$. It is easier to show the contrapositive: if $y\notin A$, then $y\notin\bar{A}$.

For $y\notin A$, there are two cases.

(1) $y(x)=a$ with $a\notin\{0,1\}$ for some $x\in I$. There exists a neighborhood (nbhd) $U_a$ of $a$ such that $U_a\cap\{0,1\}=\emptyset$. Let $U=\prod_{i\in I}X_i$ with $X_i=U_a$ for $i=x$, $X_i$ is open for finitely many $i$, and $X_i=I$ for the remaining $i$. Then $U$ is a nbhd of $y$ and $U\cap A=\emptyset$. Therefore, $y\notin\bar{A}$.

(2) $y(x)\in\{0,1\}$ for all $x\in I$ but $y(x)=1$ for more than two $x\in I$, say, $x_1$ and $x_2$, i.e., $y(x_1)=y(x_2)=1$. There exists a neighborhood (nbhd) $U_1$ of $1$ such that $0\notin U_1$. Let $U=\prod_{i\in I}X_i$ with $X_i=U_1$ for $i=x_1,x_2$, $X_i$ is open for finitely many $i$, and $X_i=I$ for the remaining $i$. Then $U$ is a nbhd of $y$ and $U\cap A=\emptyset$. Therefore, $y\notin\bar{A}$.

Answer 2

Let $X=I^I$. For $F\subseteq I$ finite and $h\in X$ let $B(F,h)=\{g\in X\mid g=h\text{ on }F\}$, which by definition of the product topology is a basic open set. Now note that the set of functions in $I^I$ that take the value $1$ at exactly one point is closed in $I^I$, being the complement of $$\bigcup_{\stackrel{\{a,b\}\in X^2}{a\neq b}}B(\{a,b\},1_{\{a,b\}}).$$

A similar argument establishes that the set of functions in $X$ that take values in $\{0,1\}$ is closed in $X$, hence the image of $f$, being the intersection of the two is also closed (and in particular Borel).

Note also that since convergence in $X$ is pointwise convergence of functions it is very easy to convince oneself that the image of $f$ is sequentially closed, and if you have familiarity with nets you can also show that it is closed in a more straightforward manner