MHB Proving A Subset B & C $\Rightarrow$ A Subset (B $\cup$ C)

  • Thread starter Thread starter cbarker1
  • Start date Start date
cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
Dear Everybody,

I am struggling now for determining if the following statements are true or false. If the statement is true, then prove it. If not, make a counterexample.
Here are the statements:
  1. $A \subset B$ and $A \subset C$ if and only if $A \subset (B\cup C)$.
  2. $A \subset B$ or $A \subset C$ if and only if $A \subset (B\cap C)$.

My attemption:
Let A={1,2,3}, B={1,2,4}, and C={3}.
  1. I believe this is true. $A\subset B$ and $A\subset C$. Thus $A \subset (B\cup C)$ is true.
  2. I don't know how to begin.
Thanks,
Cbarker1
 
Physics news on Phys.org
Hi Cbarker1.

The two statements are in fact false, thus you just need to find a counterexample to disprove each. Hints:

  1. The $\Rightarrow$ part is true but not the $\Leftarrow$ part. Consider $A\subset B$, $A\ne\emptyset$, and $C=\emptyset$.
    -
  2. This time $\Leftarrow$ is true but not $\Rightarrow$. Consider $A\subset B$, $A\ne\emptyset$, and $B\cap C=\emptyset$.
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.

Similar threads

Back
Top