Confusion with membership and subsets From what I understand, if $A$ is a set, then $B$ is a subset of $A$ if and only if all the members of $B$ are also members of $A$. $(B ⊆ A)$
However, I have come across two questions whose solutions I can't understand
$\{b,c\}⊆\{\{a,b\},\{b,c\},\{a,c\}\}$ (False) 
Why this is false when it appears that $\{b,c\}$ is contained within the set $\{\{a,b\},\{b,c\},\{a,c\}\}?$
Similiarly, I understand that membership as: $A∈B$ if $B$ is a set and A belongs to it
$\{a,b,c\} ∈ \{b,c,a\}$ (False)
I don't understand why this is false, as all the members of $A$ ($a, b$ and $c)$ appear to be members of $B$ ($b, c, a$)?
I would appreciate any help in clarifying this, thank you
 A: One must be careful to distinguish between "contains as an element" and "contains as a subset". 
An object is an element of a set of that element exists within the set. This means that the single object appears in the specification of the set, without being surrounded by braces.
A set $S$ is a subset of a given set $X$ of every element of $S$ is an element of $X$.
In your case, it is true that the set $\{b,c\}$ is an element but it is not a subset, since neither $b$ nor $c$ appear as elements of the right hand side.
A: Intuitive idea: a set is a unordered list of elements.
$x\in A$ means "$x$ is an element of $A$" ($x$ is in the list).
In your example, $\{b,c\}\in\{\{a,b\},\mathbf{\{b,c\}},\{a,c\}\}$ is true because $\{b,c\}$ is in the list.
$B\subseteq A$ means $x\in B\implies x\in A$.
In your example, $\{b,c\}\subseteq\{\{a,b\},\{b,c\},\{a,c\}\}$ is false because $b\not\in\{\{a,b\},\{b,c\},\{a,c\}\}$ and $c\not\in\{\{a,b\},\{b,c\},\{a,c\}\}$ (failing in one case is enough).
A: $\{b,c\}$ is not a subset of $\{\{a,b\},\{b,c\},\{a,c\}\}$ because $\{\{a,b\},\{b,c\},\{a,c\}\}$ does not contain the elements $b$ and $c.$ In the same way, $\{b,c,a\}$ does not contain the whole set $\{a,b,c\}$ so it doesn't belong to the former set. It would be true if the statement was :
$\{\{a,b,c\}\} \in \{b,c,a,\{a,b,c\}\}$
Hope it clears your doubt.
A: $\underline{As\space to \space the \space first \space point}$. 

If a set $A$ is contained in a set $B$ ( meaning , is a subset of
  set $B$) , then $A\cup B=B$.

Intuitively, if $A$ is a subset of $B$, then ( by definition) all the elements of $A$ are also elements of $B$, and , therefore, $A$ " adds" nothing to $B$ when $A$ is "united" to $B$. 
In order to apply the test to the case under consideration, 
let  $A=\{b,c\}$ and $B= \{\{a,b\},\{b,c\},\{a,c\}\}$ . 
In that case what is $A\cup B$? 
$A\cup B$
$=$ the set of all $x$ that belong to $A$ or to $B$ ( inclusive " or")
$= \{a, b, \{a,b\},\{b,c\},\{a,c\}\}$. 

You can see easily that $A\cup B\neq B$.

For example, object $a$ belongs to $A\cup B$ but does not belong to set $B$. Same thing for object $b$. 
Supplementary reason: $B$ was a $3$ elements set, while  $A\cup B$ is a $5$ elements set. So they cannot be one and the same set. 

Since set $A$ " adds" something to set $B$ when A is "united" to $B$,
  $A$ is not a subset of B.

$\underline{As\space to \space the \space second \space point}$. 
Your reasoning is as follows : 
(1) All the elements of $A$ are also elements of $B$
(2) Therefore, $A$ should be a member of $B$. 
But the premise is irrelevant to the conclusion, since your premise is the definition of the inclusion ( i.e. subset) relation, while your conclusion is a membership claim. 
