Example of Borel set for the graph of a function mapping a real interval to a function 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 graph of $f$ is a Borel set in $I\times I^I$.

My efforts:

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

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

$\{y\times\{1_{\{y\}}\}\}$ is Borel in $I\times I^I$ for any $y\in I$.

Countable union of Borel sets is Borel. Thus $\{y\times1_{\{y\}}:y\;\mathrm{is\;rational},y\in I\}$ is Borel in $I\times I^I$.

Then I don't know how to continue.

Answer 1

Let $G=\{\langle\alpha,f(\alpha)\rangle:\alpha\in I\}$. Suppose that $\langle\alpha,y\rangle\in\left(I\times I^I\right)\setminus G$.

• If $y_\xi\in(0,1)$ for some $\xi\in I$, then $$I\times\left\{z\in I^I:z_\xi\in(0,1)\right\}$$ is an open nbhd of $\langle\alpha,y\rangle$ disjoint from $G$.
• If there is a $\xi\in I\setminus\{\alpha\}$ such that $y_\xi=1$, then $$(I\setminus\{\xi\})\times\left\{z\in I^I:z_\xi\in(0,1]\right\}$$ is an open nbhd of $\langle\alpha,y\rangle$ disjoint from $G$.

The only remaining possibility is that $y=1_\varnothing$, the zero function on $I$, and in that case $\langle\alpha,y\rangle\in\operatorname{cl}G$, since every non-empty open set in $I$ is infinite. Thus,

$$\operatorname{cl}G=G\cup(I\times\{1_\varnothing\})\,,$$

and $G=(\operatorname{cl}G)\setminus(I\times\{1_\varnothing\})$. $G$ is therefore the difference of two closed sets and as such is Borel.