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When asked to give the powerset of the set $a=\{3,5\},$ would I also include the set $(a)$ within the powerset.

So this would be $p(a) = \{\phi, \{3\}, \{5\}, \{3,5\} \}$

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    Yes, of course, since $a=\{3,5\}$ is a subset of $a.$ – 2017-01-03
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    Yes, of course. – 2017-01-03
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    @Rohan I wouldn't use the word "obviously" here. If it was obvious to OP, they wouldn't ask, and using that kind of language comes across as a bit passive aggressive. – 2017-01-03
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    Alright here's a good time for a pointer regarding the use of this site: **downvotes on questions should be for ill-posed or legitimately bad questions, not posts you think are beneath you or this site**. This voting behavior is toxic to the long term growth of this site. This question is perfectly fine and OP even has an answer in their post. (Not to mention the downvotes on the perfectly fine answers.) – 2017-01-03
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    @CameronWilliams I love you. – 2017-01-03
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    @CameronWilliams, by using the wording "of course", we are not offending the OP, but we are using the tools the language gives us to note that the answer does not rely on a complicated mathematical matter, but on a basic remark regarding sets. – 2017-01-03
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    @NikolaosSkout it didn't say "of course" before. It said "obviously" which is a fairly loaded word. – 2017-01-04

3 Answers 3

3

Yes. The power set of $a$ is the set of all subsets of $a$, and every set is a subset of itself.

2

Yes. That is the correct power set for $a$.

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    Thanks @Nathan H. Someone has decided to edit my message so just to clarify if i have a set a = {3,5} then the powerset of this is p(a) = {∅, {3}, {5} {3,5}, a} OR p(a) = {∅, {3}, {5} {3,5}} – 2017-01-03
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    The second one. (Though technically, both of those since those are both the same set. Sets can't have duplicates so for example the set $\{1,1\}$ is equal to the set $\{1\}$.) – 2017-01-03
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Yes, the power set of a given by you is correct. Actually the thing which is making you confusing is that every set is subset of its own.

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    I have the very strong feeling that you have at least three distinct accounts and you upvote yourself using them. For instance, in this post three are three answers, your answer is the newest one and has three upvotes and it's pretty much the same as the other ones so I see no reason for the 2 extra upvotes. Also, in [here](http://math.stackexchange.com/questions/2070170/solving-2x42x2y2y4-z2/2070767#2070767) there is simply no reason for the upvotes to your "answer" (it had 3 upvotes when I first saw it and the sole fact that it has upvotes is just ridiculous)... – 2017-01-03