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I tried looking at some of the questions that could apply, but I'm not sure that they applied to this type of problem.

Given $A \in \mathcal{P}(X)$ define the characteristic function $ \mathcal{X}_{A}:X\rightarrow \{0,1\}$ by $$\mathcal{X}_A(x) = \begin{cases}0 & if ~ x\notin A, \\ 1 & if ~ x \in A. \end{cases}$$

Suppose that A and B are subsets of X. Prove that the function $x\rightarrow \mathcal{X}_A(x)\mathcal{X}_B(x)$ (multiplication of integers) is the characteristic function of the intersection $A \cap B$.

So, as far as I understand, then A is an element in the power function of X and A and B are subsets of X, but I'm not sure how to get started... Any help?

Daniel Cazares
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2 Answers2

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Suppose that $x \in A \cap B$; then $\chi_A(x) = 1$ and $\chi_B(x) = 1$ implies that $\chi_A (x) \chi_B(x) = 1$.

Now if $x \notin A \cap B$, can you show that one of $\chi_A(x)$ and $\chi_B(x)$ is zero? This will prove that

$$\chi_A(x) \chi_B(x) = \left\{\begin{array}{r} 1 & x \in A \cap B \\ 0 & \text{ else}\end{array}\right.$$

which is exactly what you're trying to prove.

  • Thanks! For the second part, would it be appropriate to say "Without loss of generality, suppose $x\in A$ and $x\notin B$." – Daniel Cazares Nov 15 '13 at 20:43
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    @thiefjack Not really, because you could have $x \notin A$ and $x \notin B$; what you could say is that without loss of generality, $x \notin A$. –  Nov 15 '13 at 20:47
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$ \newcommand{ifelse}[3]{(#1\text{ if } #2\text{ else } #3)} $Here is how I would prove this: through calculation. Starting with the most complex side, expand the definitions and try to simplify: for all sets $\;A,B\;$, for every $\;x\;$, and occasionally taking baby steps, we have \begin{align} & \chi_A(x) \times \chi_B(x) \\ = & \qquad \text{"definition of $\;\chi\;$, twice"} \\ & \ifelse{1}{x \in A}{0} \times \ifelse{1}{x \in B}{0} \\ = & \qquad \text{"$\;\times\;$ distributes over $\;\ifelse{}{}{}\;$"} \\ & \ifelse{\ifelse{1}{x \in A}{0} \times 1}{x \in B}{\ifelse{1}{x \in A}{0} \times 0} \\ = & \qquad \text{"arithmetic: simplify"} \\ & \ifelse{\ifelse{1}{x \in A}{0}}{x \in B}{0} \\ = & \qquad \text{"logic: simplify"} \\ & \ifelse{1}{x \in A \land x \in B}{0} \\ = & \qquad \text{"definition of $\;\cap\;$"} \\ & \ifelse{1}{x \in A \cap B}{0} \\ = & \qquad \text{"definition of $\;\chi\;$"} \\ & \chi_{A \cap B}(x) \\ \end{align} This completes the proof.

MarnixKlooster ReinstateMonica
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