0
$\begingroup$

In How do I prove $A \cup\varnothing = A$ and $A \cap\varnothing = \varnothing$

A proof was given reproduced here:

Prove: $A \cup \varnothing = A$

Let $a\in A\cup \varnothing$. Then $a\in A$ or $a\in\varnothing$. Since $a\in\varnothing$ is false regardless of $a$, but we assumed $a\in A\cup \varnothing$, it must be that $a\in A$ is true, so that $A\cup \varnothing \subseteq A$. Conversely, $A\cup \varnothing \supseteq A$ trivially, so $A=A\cup\varnothing$.

I never took a course on elementary logic (or real analysis for that matters) so it escapes me:

it must be that $a\in A$ is true $\implies A\cup \varnothing \subseteq A$

How does a truth statement (a sentence in Englisch) just translates into a set inclusion??

Should there be something in between:

it must be that $a\in A$ is true [and in logic, "true" relates to set inclusion like this] $ A\cup \varnothing \subseteq A$

Can someone bridge this gap?

$\endgroup$
0

3 Answers 3

Reset to default
4
$\begingroup$

Here is how the full logic of the proof goes:

  1. $a \in A \cup \emptyset$

  2. $a \in A \cup \emptyset \rightarrow (a \in A ~\vee~ a \in \emptyset)$ by the definition of the union of two sets.

  3. $(a \in A ~\vee~ a \in \emptyset)$ by Modus ponens.

  4. $a \notin \emptyset$ by the definition of the empty set as the set with no elements.

  5. $(a \in A ~\vee~ a \in \emptyset) ~\wedge~ (a \notin \emptyset) \rightarrow a \in A $ by disjunctive syllogism.

  6. $a \in A$ by Modus ponens

  7. $a \in A \cup \emptyset \rightarrow a \in A$ by the transitive property of the conditional operator.

  8. $\therefore ~ A \cup \emptyset \subseteq A$ by the definition of a subset.

$\endgroup$
5
  • $\begingroup$ Thank you, this is what I was looking for. What gives us "$a \in A \cup \emptyset \rightarrow (a \in A ~\vee~ a \in \emptyset)$"? I know it is obvious but would be the proof for this? The left hand side has no logic operator, how does the logic operator $\lor$ just appear on the right hand side? $\endgroup$
    – Shamisen Expert
    Mar 25, 2016 at 4:15
  • 1
    $\begingroup$ We define $A \cup B$ to be equal to the set $\{x|x \in A \vee x \in B\}$, so if some element $a$ is in the union of two sets, then it must be in one of those two sets. $\endgroup$
    – TreeHouse196
    Mar 25, 2016 at 4:17
  • $\begingroup$ Thanks, I didn't know the union had a definition. In engineering we just use proof by intuition. It works until someone asks you to give a formal proof, then you crash $\endgroup$
    – Shamisen Expert
    Mar 25, 2016 at 4:18
  • $\begingroup$ Also can you add a simple intuition to this "disjunctive syllogism" thing? I looked up on Wikipedia and it seems pretty wild $\endgroup$
    – Shamisen Expert
    Mar 25, 2016 at 4:49
  • $\begingroup$ Disjunctive syllogism says that if P or Q is true, and Q is false, then P must be true. P or Q is true if and only if P is true, Q is true, or P and Q are both true. If Q is false, then the only way P or Q can be true is if P is true. $\endgroup$
    – TreeHouse196
    Mar 25, 2016 at 4:50
1
$\begingroup$

Definition of $B\subseteq A$ is $\forall x\in B:x\in A$
Substitute $B=A\cup\emptyset$ and you will get it.

$\endgroup$
0
$\begingroup$

The statement "Let $a \in A \cup \emptyset$" says that $a$ can be any element of $A \cup \emptyset$. So by showing that $a \in A$, we have shown that any element of $A \cup \emptyset$ is an element of $A$, or equivalently that $A \cup \emptyset \subset A$ (this is in fact the definition of subset).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.