Question

what is wrong with the following proof?
If A \subseteq A\cup B, then either A \subseteq B or A \subseteq C.

Proof: suppose not, then A \nsubseteq B and A \nsubseteq C. Suppose x \in A, then x \notin A and x \notin B. In other words, ~ (x \in A or x \in B). Which means x \notin A \cup B. Which means A \notsubseteq B \cup C

Answer 1

damn you LaTex!!

Answer 2

heheh

Answer 3

don't forget the \( part

Answer 4

and don't forget to close \)

Answer 5

hold on ...

Answer 6

counter example is
A = {1,2} , B = {1}, C = {2}

Answer 7

u hae p-->Q
suppose not , so u wanna suppose ~Q right ?

Answer 8

yeah, I was using contrapositve. Isn't this the same as contradiction?

Answer 9

no , contrapositive
is like this
~Q-->~p

Answer 10

I have typos. Hold on,
if A ⊆ B∪C, then either A⊆B or A⊆C

proof: Suppose not, then
A not ⊆ B and A not ⊆ C. Let x ∈ A, then x∉B and x∉C. In other words,
~(x∈B or x∈C). Which is, ~ (x ∈ B∪C), which is A not ⊆ B∪C. *contradiction*

Answer 11

Sorry for my typos XD. Hope i'm not confusing you guys

Answer 12

so u wanna suppose ~( A⊆B or A⊆C) ?

Answer 13

yeah

Answer 14

~( A⊆B or A⊆C)=A⊆~B and A⊆~C

Answer 15

by ⊆~, you mean "not subset" right ?

Answer 16

so A⊆~B U~C

Answer 17

oh , A is a group not an elemant ops !

Answer 18

yeah, A,B,C are sets

Answer 19

well , ur prove is correct i cant see somthing wrong with it :|

Answer 20

No, the proof isn't correct. I gave an counter example above

Answer 21

but this sentense is messing somthing
if A ⊆ B∪C, then either A⊆B or A⊆C
i feel its wrong

Answer 22

Well, I copied right out of the text book.

it literally said:
"Theormem? if A ⊆ B∪C, then either A⊆B or A⊆C"

So it wants me to either prove or disprove.

Answer 23

so its clear , just give a counter exampl, dnt need a prove

Answer 24

Well, my goal is to figure out why the proof above is invalid.

Answer 25

becuz , its wrong sentence
P-->Q givs u F

Answer 26

I think I might have found my mistake here.
A not subset B does not necessarily means ALL elements in A are not in B (which is what I said). It only means there exist some A that are not in B

Answer 27

but i think u cant prove somthing wrong !
so if u have a counter example then lol
u dnt need to disprove using some mathmetics methods
it wont be right

Answer 28

hehe yeah. Why the heck am i torturing myself XD

Answer 29

unless u proved this sentence
\[if A ⊆ B \cup C, then A⊆B \and A⊆C\]

Answer 30

$$
x \in A \text{ does not imply that } x \notin B
$$

Let P represent the statement A ⊆ B∪C
Let Q represent the statement A⊆B or A⊆C

We then want to prove that \(NOT(Q) \rightarrow NOT(P)\) by contrapositive, which is equivalent to proving that \(P\rightarrow Q\).

The negation of Q is
$$
NOT(x\in A \rightarrow x\in B\cup x\in A \rightarrow x\in C)\\
\iff NOT(x\notin A \cup ~x\in B) \cap NOT(x\notin A\cup ~x\in C)\\
\iff (x\in A\cap~x\notin B)\cap(x\in A\cap~x\notin C)\\
\iff (x\in A \cap x \notin B \cap x \notin C)\\
\iff NOT(x\notin A\cup (x\in B\cup x\in C))\\
\iff NOT(x\in A \rightarrow (x\in B\cup x\in C))\\
\iff NOT(A ⊆ B∪C)\\
\equiv NOT(P)
$$

Therefore, if A ⊆ B∪C, then either A⊆B or A⊆C, is true.