Question

Prove that A ∪ ∅=A

Answer 1

how rigorous of a proof is what i'm wondering...

Answer 2

Hmm.. me too.

Answer 3

this is the venn diagram of a union \(\phi\)

Answer 4

\[\begin{align}A \cup \emptyset &= \{x|x\in A \wedge x\in\emptyset\}\\
&=\{x|x\in A\}\\
&=A
\end{align}\]
since \(x\in\emptyset=F\), so that \(x\in A=T\).

Answer 5

but \(\phi\) is nothing, zilch! so anything combined with zilch is that thing itself!!

Answer 6

right @apoorvk hence proved :)

Answer 7

phi is the empty set

Answer 8

you prove two sets are equal by showing containment in each direction, i.e. showing \
\[A\subseteq (A \cup \emptyset)\] and
\[(A \cup \emptyset)\subseteq \cup A\]

Answer 9

Btw, HAPPY BIRTH CENTENARY to ALAN TURING, aka @TuringTest in this life :P

Answer 10

Hence, Proved -_- @mathslover lol

Answer 11

lol

Answer 12

or you can use blockcolder way but i think there is a typo there
\[\begin{align}A \cup \emptyset &= \{x|x\in A \lor x\in\emptyset\}\\
&=\{x|x\in A\}\\
&=A
\end{align}\]

Answer 13

Thanks for all your responses .

Answer 14

Oh yeah. Never noticed that. Thanks for the correction!

Answer 15

@Hollywood_chrissy you can do this the slow way by picking \(x\in A\) and showing it is in \(A\cup \emptyset\) which it is, since it is in A and then pick \(x\in A\cup \emptyset\) and showing it is in A, since \(\emptyset\) is empty, if \(x\in A\cup \emptyset\) then \(x\in A\) and you are done

Answer 16

suppose \(A=\mathbb{R}\)?

Answer 17

\[A=\{a_0, a_1,a_2,\cdots,a_n\}\qquad ø=\{\}\]
\[A+ø=\{a_0,\}+\{a_1\}+\{a_2\}+\cdots+\{a_n\}+\{\}\]\[\qquad=\{a_0, a_1,a_2,\cdots,a_n\}=A\]