Question 476252
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".

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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 
 
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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.

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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

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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.

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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.

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So only A ⊂ C of the above statements is false.


Edwin</pre>