I'm working on some tasks which is listed below, and I'm trying to figure out if I've understood partition, power set, and complements correctly.

Here are the tasks:

Assume that $\{$$1, 2, 3, 4, 5$$\}$

  1. What is the complement of the amount $\{$$1, 2, 3$$\}$?

  2. What is $P(${$1, 4$}$)$ ie power set of the amount {$1, 4$$\}$?

  3. Give a partition of the amount {$a, b, c, d, e, f$} so that {$a, b, c, d$} is included

  4. Give a partition of the amount {$1, 2, 3, 4$} which has two elements

  5. Why isn't $\{$$\{$$1, 2$$\}$, $\{$$2, 3$$\}$$\}$ a partition of {$1, 2, 3$$\}$?

  6. Is it always the case that $X \in P(X)$, no matter what the $X$ is?

Here is my answers:

  1. $\{$4, 5$\}$

  2. $\{$$\emptyset$, $\{$1$\}$, $\{$4$\}$, $\{$1, 4$\}$$\}$

  3. Not sure

  4. Not sure

  5. Not sure

  6. Not sure

I would appreciate If someone could go through the tasks which I've done and see if they are correct. Would also appreciate help with $(3)$, $(4)$, $(5)$ and $(6)$.

Thank you.

  • $\begingroup$ I think this question is not related to the first-order logic. $\endgroup$ – Hanul Jeon Nov 11 '13 at 15:26
  • $\begingroup$ hmm, what is it related to then, so I can edit $\endgroup$ – Dabbish Nov 11 '13 at 15:26

The first two are correct. The wording in (3) is very strange — I suspect that you’re translating — but I think that it’s asking for a partition of the set $\{a,b,c,d,e,f\}$ that has $\{a,b,c,d\}$ as one of its parts. There are two such partitions: $$\big\{\{a,b,c,d\},\{e,f\}\big\}\;,$$ and $$\big\{\{a,b,c,d\},\{e\},\{f\}\big\}\;.$$ Either is a correct answer; the crucial thing is that each member of $\{a,b,c,d,e,f\}$ must appear in exactly one member of the partition.

There are lots of two-element partitions of $\{1,2,3,4\}$; here are just a few of them:

$$\begin{align*} &\big\{\{1\},\{2,3,4\}\big\}\\ &\big\{\{1,2\},\{3,4\}\big\}\\ &\big\{\{1,2,3\},\{4\}\big\}\\ &\big\{\{1,3,4\},\{2\}\big\} \end{align*}$$

A partition of $\{1,2,3\}$ is a collection of non-empty subsets of $\{1,2,3\}$ such that each member of $\{1,2,3\}$ is in exactly one of the subsets. $\big\{\{1,2\},\{2,3\}\big\}$ is not a partition of $\{1,2,3\}$ because the element $2$ belongs to two members of the collection: $2\in\{1,2\}$ and $2\in\{2,3\}$. Another way to say it is that the members of a partition must be disjoint from one another, and $\{1,2\}$ and $\{2,3\}$ are not disjoint: they have $2$ in common.

The answer to the last one is yes: if $X$ is any set, then $X\subseteq X$, so by definition $X\in\wp(X)$.

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  1. good

  2. good

  3. $\{\{a,b,c,d\},\{e,f\}\}$ or $\{\{a,b,c,d\},\{e\},\{f\}\}$

  4. $\{\{1,2\},\{3,4\}\}$

  5. The element 2 occurs in both partitions, hence they do not have non-empty intersection.

  6. Yes. Can you see why?

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  • $\begingroup$ regarding the typo on $(3)$ I edited the post. I think it's supposed to say "that has one of the parts" or "included" at the end, if that makes any sense. $\endgroup$ – Dabbish Nov 11 '13 at 15:34

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