True or false question

If B is a subset of A then {B} is an element of power set A.

I think this is true.

Because B is {1,2} say A {1,2,3} then power set of includes


Unless {B} means $\{\{1,2\}\}$

  • 2
    $\begingroup$ That is what $\{B\}$ means, $\{\{1,2\}\}$. So the answer is false; if $B$ is a subset of $A$ then $B$ (not $\{B\}$) is an element of $\mathcal{P}(A)$, for this is the definition of the power set. However, you can create sets in which there is a subset of $A$ that is also an element of $A$, so more appropriately the answer should be 'not necessarily'. $\endgroup$ – Hayden Jun 18 '14 at 17:20

It is true that $B$ is in the power set of a set $A$ (we'll call the powerset $P(A)$) is the set of all subsets of $A$, so the elements of $P(A)$ include all subsets of $A$.

Since $B$ is given to be a subset of $A$, then it is an element in the powerset of $A$.

However, $\{B\}$ is a set containing the subset $B$ as its only element, and so $\{B\}$ is not in the powerset of $A$.

  • $\begingroup$ But $\{B\}$ is not, in general. $\endgroup$ – Stefan Mesken Jun 18 '14 at 17:25
  • $\begingroup$ I see so B is a subset of A, then B itself is an element of power set A,......, I see hmm oh yes a tricky question because {B} does not equal B,...., set theory should be called bracket theory $\endgroup$ – Fernando Martinez Jun 18 '14 at 17:27
  • $\begingroup$ I like that: "Bracket Theory" :-) $\endgroup$ – Namaste Jun 18 '14 at 17:33
  • $\begingroup$ Technically, we can't say for sure that $\{B\}$ is not in the power set of $A$ since we don't know whether or not $B$ is an element of $A$. It depends on the sets in question. $\endgroup$ – Santiago Canez Jun 18 '14 at 18:11

The definition of a power set of a set $A$ is the set of all subsets of $A$ including $A$ itself and the null set. As $B$ is a subset of $A$ in your question, then yes $B$ is an element in the power set of $A$.

  • 1
    $\begingroup$ This doesn't answer the question as to whether $\{B\}$ is an element of the power set. $\endgroup$ – Santiago Canez Jun 18 '14 at 18:07
  • $\begingroup$ You are correct; I missed that subtlety which amWhy explained. $\endgroup$ – Avraham Jun 18 '14 at 18:34

I think it is true... Because the in asked question the {B} is a power set of A. It meabs all the elements of B are exist in A as Set B us the subset of set A. So, the power set of A will also contain all the sub sets of Set B.


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