Question 476252: 6. Given the following sets, select the statement below that is NOT true.
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
(Points : 2)
B ⊂ A
A ⊂ C
B ⊆ A
C ⊆ A
C ⊆ B
thank you for your help.
Answer by Edwin McCravy(20059)   (Show Source):
You can put this solution on YOUR website! Note the difference between "subset" and "proper subset".
⊆ and ⊂. Both symbols require that every member of the set
on the left of the symbol is also a member of the element on
the right of the symbol. However with ⊂, the set on the left
cannot contain ALL OF the elements of the set on the right of ⊂.
IOW, ⊆ and ⊂ are the same EXCEPT when the same set is on both
side of the symbol. "A ⊆ A" is true but "A ⊂ A" is false.
⊆ allows (but does not require) that the set on the left of ⊆
contain ALL the members of the set on the right of ⊆. The
symbol ⊂ does not allow that the set on the left contain ALL
the elements of the set on the right of it. IOW,
{t,o,p} ⊆ {p,o,t} is true
but
{t,o,p} ⊂ {p,o,t} is false
To put it another way:
The only "improper subset" is the set itself, because it's
not properly "sub" which means "less than".
---------------------------------------------------------
B ⊂ A
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is true because every member of of B = {r, i, s, e}
is also a member of A = {r, i, s, k, e, d} yet B is not
equal to A. So B is a PROPER subset of A
----------------------------------------------------------
A ⊂ C
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is false because not every member of of
A = {r, i, s, k, e, d} is a member of C = {s, i, r}, since
C has no k, e, or d. So B is a subset of A.
-----------------------------------------------------------
B ⊆ A
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is true because every member of of B = {r, i, s, e}
is also a member of A = {r, i, s, k, e, d}. So B is a PROPER
subset of A. B doesn't contain ALL of A
-----------------------------------------------------------
C ⊆ A
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is true because every member of of C = {s, i, r}
is also a member of A = {r, i, s, k, e, d}. So C is a PROPER
subset of A.
-----------------------------------------------------------
C ⊆ B
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is true because every member of of C = {s, i, r}
is also a member of B = {r, i, s, e}. So C is a PROPER subset
of B.
----------------------------------------------------------
So only A ⊂ C of the above statements is false.
Edwin
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