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?


3 Answers 3


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.

  • $\begingroup$ 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? $\endgroup$ 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. $\endgroup$ 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 $\endgroup$ 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 $\endgroup$ 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. $\endgroup$ Mar 25, 2016 at 4:50

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


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).


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