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I am reading the book "A Transition to Advanced Mathematics" 2014 edition, written by Smith, Eggen and St. Andre.

For every set X:

∅ ⊆ X

∅ ∈ power set of X

∅ ⊆ power set of X

{∅} ⊆ power set of X

Is {∅} ⊆ power set of X wrong?

I am an undergraduate student. Please help me!

The definition of power set of X is the set whose elements are subsets of X. (page 90 of the book)

I have this doubt because I remember that I read somewhere that ∅ ≠ {∅} because a set with the empty set is not empty.

Thanks for your answers!

What about this question:

If the definition of power set of X is the set whose elements are subsets of X, and {∅} is not a subset of X. How {∅} can be a subset of the power set of X? Is not that an contradiction?

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  • $\begingroup$ Everything you wrote is correct. What is your question? $\endgroup$
    – MJD
    Feb 15, 2015 at 23:35
  • $\begingroup$ @MJD: "Is {∅} ⊆ power set of X wrong?" $\endgroup$
    – Asaf Karagila
    Feb 15, 2015 at 23:38
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    $\begingroup$ Note that if $A$ is a set, then $x \in A$ iff $\{x\} \subset A$. $\endgroup$
    – copper.hat
    Feb 15, 2015 at 23:46
  • $\begingroup$ The empty set is either ∅ or {}. Empty set = {∅} is wrong. $\endgroup$ Jul 27, 2019 at 18:28

4 Answers 4

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Recall that $A\subseteq B$ if whenever $a\in A$, then we have that $a\in B$ as well.

Since $\varnothing\in\mathcal P(X)$, it means that every element of $\{\varnothing\}$ is also an element of $\mathcal P(X)$, so $\{\varnothing\}\subseteq\mathcal P(X)$ is indeed correct.

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  • $\begingroup$ Thanks for your answer! In your perspective ∅ = {∅}? $\endgroup$
    – Beginner
    Feb 15, 2015 at 23:44
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    $\begingroup$ @Beginner: Absolutely not! $\varnothing$ is a set with no elements, while the set $\{\varnothing\}$ is a set with one element (namely, $\varnothing$). $\endgroup$ Feb 15, 2015 at 23:47
  • $\begingroup$ @BrianM.Scott If the definition of power set of X is the set whose elements are subsets of X, and {∅} is not a subset of X. How {∅} can be a subset of the power set of X? Is not that an contradiction? $\endgroup$
    – Beginner
    Feb 16, 2015 at 0:02
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    $\begingroup$ @Beginner: No contradiction. $\varnothing$ is a subset of $X$, so $\varnothing$ is an element of $\wp(X)$, just as you said. And since $\varnothing$ is an element of $\wp(X)$, automatically $\{\varnothing\}$ is a subset of $\wp(X)$. This is no different from observing that $3\in\Bbb N$, so $\{3\}\subseteq\Bbb N$. $\endgroup$ Feb 16, 2015 at 0:06
  • $\begingroup$ I think that I am starting to understand! Thanks! $\endgroup$
    – Beginner
    Feb 16, 2015 at 0:07
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The empty set $\emptyset$ is a subset of every set $X$. Hence $\emptyset$ is a member of the power set of $X$, $P(X)$. Therefore $\{\emptyset\}$ is a subset of $P(X)$.

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  • $\begingroup$ Thanks for your answer! In your perspective ∅ = {∅}? $\endgroup$
    – Beginner
    Feb 15, 2015 at 23:45
  • $\begingroup$ No, that is false. See comments above. $\endgroup$
    – Simon S
    Feb 15, 2015 at 23:57
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Let's do this with an example.

Here's a set: $\{0,1\}$. Its power set ($\mathcal{P}$) is: $\{ \emptyset,\{0\},\{1\},\{0,1\} \}$. That's four elements. We can check that we calculated the power set correctly because we know that for a set $A$ with size $|A|$, the size of the power set is always $2^{|A|}$, and $2^2$ is $4$.

Here's another example: take the set $\{2,3,4\}$. It is clear that $\{2,3\}$ is a subset. And so $\{2\}$ is a subset as well.

With this example, we can see that $\{\emptyset\} \subseteq \mathcal{P}$ holds. Clearly $\{\emptyset,\{0\}\}$ is a subset of $\mathcal{P}$, so $\{\emptyset\}$ must be, too, even if may look unusual at a first glance.

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  • $\begingroup$ Thanks for your answer! In your perspective ∅ = {∅}? $\endgroup$
    – Beginner
    Feb 15, 2015 at 23:45
  • $\begingroup$ @Beginner You're welcome, and no. Read it out to yourself: one of these is the empty set. The other is the set containing the empty set. These are different objects. $\endgroup$
    – Newb
    Feb 15, 2015 at 23:46
  • $\begingroup$ Note that $\emptyset$ is a set containing zero elements whereas $\{\emptyset\}$ is a set containing one element. $\endgroup$
    – copper.hat
    Feb 15, 2015 at 23:48
  • $\begingroup$ @copper.hat Thanks for your help! I am clear now. $\endgroup$
    – Beginner
    Feb 16, 2015 at 0:11
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One of those statements doesn't belong with the others. "∅ ⊆ power set of X" is correct, but irrelevant. I think that's what's confusing you.

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