Given $A = \{1, 2\}$ and $B = \mathbb{R}$, we need to sketch the graph of $R = A \times B$. We need to determine if $R$ is a function from $A$ to $B$. If it is not, we need to give a subset of $R$ which is a function.

1. Problem Description

Given $A = \{1, 2\}$ and $B = \mathbb{R}$, we need to sketch the graph of $R = A \times B$. We need to determine if $R$ is a function from $A$ to $B$. If it is not, we need to give a subset of $R$ which is a function.

2. Solution Steps

First, we find the Cartesian product $R = A \times B$. Since $A = \{1, 2\}$ and $B = \mathbb{R}$,

$R = A \times B = \{(a, b) \mid a \in A, b \in B\} = \{(1, b) \mid b \in \mathbb{R}\} \cup \{(2, b) \mid b \in \mathbb{R}\}$.

The graph of $R$ consists of two vertical lines, $x = 1$ and $x = 2$, on the $xy$-plane.

To determine if $R$ is a function from $A$ to $B$, we need to check if each element in $A$ is mapped to exactly one element in $B$. In other words, for each $a \in A$, there should be a unique $b \in B$ such that $(a, b) \in R$.

In this case, $R$ is not a function from $A$ to $B$. For example, $1 \in A$ is mapped to infinitely many elements in $B$, namely all real numbers. So, $R$ is a relation from $A$ to $B$, but it is not a function.

Now we need to find a subset $f$ of $R$ that is a function from $A$ to $B$. A possible function $f$ can map $1$ to $0$ and $2$ to $1$. So, $f = \{(1, 0), (2, 1)\}$. Another function could be $f = \{(1, 5), (2, -3)\}$.

In general, we can define a function $f: A \to B$ such that $f = \{(1, b_1), (2, b_2)\}$ where $b_1, b_2 \in \mathbb{R}$.

Let's choose $b_1 = 0$ and $b_2 = 0$. Then $f = \{(1, 0), (2, 0)\}$.

This is a valid function because each element in $A$ maps to exactly one element in $B$.

3. Final Answer

The graph of $R$ consists of two vertical lines $x=1$ and $x=2$.

$R$ is not a function from $A$ to $B$.

A subset $f$ of $R$ which is a function is $f = \{(1, 0), (2, 0)\}$.