1
$\begingroup$

I'm reading Guerino Mazzolla's Comprehensive Mathematics for Computer Scientists 1:

Axiom 3 (Axiom of Union) If $a$ is a set, then there is a set

$$\{x| \text{ there exists an element } b \in a \text{ such that } x\in b \}$$

This set is denoted by $\cup a$ and is called the union of $a$.

$ $

Lemma 1 For any set $a$, there is the set $a^+=a\cup \{a\}$. It's called the successor of $a$.

$\emptyset^+=\emptyset \cup \{ \emptyset \}$ can be rewritten as $\bigcup \{\emptyset, \{\emptyset \} \}$, then applying the axiom of union, there are two elements $b\in a$:

  • $\emptyset$ and $\{\emptyset \}$

And then when applying $x\in b$, I'm having trouble because the book states that the empty set is the set with no elements inside it. But somehow it seems reasonable that $\emptyset \in \emptyset$, then I could proceed:

  • $\emptyset^+=\{\emptyset,\{\emptyset \} \}$

  • $\{\emptyset,\{\emptyset \} \}^+= \bigcup \{\{\emptyset,\{\emptyset \} \},\{\{\emptyset,\{\emptyset \} \} \} \}= \{\emptyset,\{\emptyset \},\{\emptyset,\{\emptyset \} \} \}$

I've also made an experiment with Mathematica that seems to confirm my hypothesis:

a = {}
a \[Union] {a}
(a \[Union] {a}) \[Union] {a \[Union] {a}}
((a \[Union] {a}) \[Union] {a \[Union] {a}}) \[Union] {(a \[Union] {a}) \[Union] {a \[Union] {a}}}

Which yields:

{}
{{}}
{{}, {{}}}
{{}, {{}}, {{}, {{}}}}
$\endgroup$
  • 1
    $\begingroup$ Your experiment disagrees with your hypothesis, since it (correctly) determines that $\{\}^+ = \{\{\}\}$, not $\{\{\}, \{\{\}\}\}$ as you hypothesized. $\endgroup$ – user14972 Jan 26 '14 at 3:42
  • $\begingroup$ Yes. Sorry, my mind was too burned for me to notice the difference. $\endgroup$ – Billy Rubina Jan 26 '14 at 4:05
  • 1
    $\begingroup$ "But somehow it seems reasonable that $\emptyset\in\emptyset$": this should not seem reasonable at all; try to formulate why you think it should be. Anyway see this answer, and more generally several questions that should show up on "Related" links on the right margin of this question. $\endgroup$ – Marc van Leeuwen Jan 26 '14 at 4:16
  • $\begingroup$ Commenting on the text: I would think that a successor function defined along the lines of Peano's axioms would be more fruitful in a CS text. It would also avoid wonky results like $2\subset 5$. $\endgroup$ – Dan Christensen Jan 27 '14 at 16:16
  • $\begingroup$ @DanChristensen The book has the Peano Axioms, but at this elementary level (in the book), the natural numbers still don't exist. The author is in the process of construction. $\endgroup$ – Billy Rubina Jan 28 '14 at 2:29
1
$\begingroup$

What is wrong with $\emptyset$ having no elements? It means that $\emptyset \cup \{\emptyset\} = \{\emptyset\} = \{\{\}\}$, which is not what you wrote, because $\{\emptyset,\{\emptyset\}\} = \{\{\},\{\{\}\}\}$.

$\endgroup$
  • $\begingroup$ Thanks. Yes, now I see. I was afraid of doing the operation and having no element to put in there. I thought that the doing the operation and picking $x\in b$ would give me an indefinite expression such as divison by $0$. $\endgroup$ – Billy Rubina Jan 26 '14 at 4:04
  • 1
    $\begingroup$ You probably should avoid thinking of it as "picking one", because there could be none. Rather, think of it as "for any set in the union (there could be no set at all), its members (there could be none) are in the union, and for any element (possibly none) in the union, it is a member of some (at least one) set in the union." $\endgroup$ – user21820 Jan 26 '14 at 4:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.