2

$\{X \in B\} \Delta \{Y \in B\} \subset \{X \ne Y\}$ was shown here.

How do I show the contrapositive (directly) ?

The contrapositive: $(\{X \in B\} \cap \{Y \in B\}) \cup (\{X \notin B\} \cap \{Y \notin B\}) \supset \{X = Y\}$

WOLOG: If $\omega \in \{\omega | X = Y\}$ and $\omega \in \{\omega | X \in B\}$, how can I say $Y \in B$?

This seems too simple to be right.

$\omega \in \{\omega | X(\omega) = Y(\omega)\}$ and $\omega \in \{\omega | X(\omega) \in B\}$

$\to \omega \in \{\omega | X(\omega) = Y(\omega)\} \cap \{\omega | X(\omega) \in B\}$

$\to \omega \in \{\omega | X(\omega) = Y(\omega) \cap X(\omega) \in B\}$

$\to \omega \in \{\omega | Y(\omega) \in B\}$...allowed because $X(\omega), Y(\omega) \in \mathbb{R}$ ?

Is that it? Am I missing something?

Community
  • 1
BCLC
  • 14,197

1 Answers1

1

It's correct, but it can be shortened a bit.

If $\omega \in \{X = Y\}$ and $\omega \in \{X \in B\}$, then $Y(\omega) = X(\omega) \in B$. Therefore $\omega \in \{Y \in B\}$. Note that this only shows the inclusion $\{X = Y\} \cap \{X \in B\} \subset \{Y \in B\}$.

The equality $\{X \in B\} \Delta \{Y \in B\} = \{X \ne Y\}$ does not hold in general. The left-hand side can be empty [e.g. if $B = \emptyset$ or $B = \mathbb{R}$], but it is easy to find examples where the right-hand side is nonempty.

Dominik
  • 20,241