# Help needed with complements, partition and power sets

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.

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?

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.

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

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?

• 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. – Dabbish Nov 11 '13 at 15:34