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I'm having issues getting my head around cartesian products and their cardinalities.

$A = \{0, 1, \{2, 3, 4\}\}$
$B = \{1,5\}$
$D = B \times N$ (where $N$ is the set of natural numbers)

The first problem: What is the cardinality of:

(a) $A \times B$ (cartesian product)

(b) $A \times D$

Part 2: true/false (a) $N$ is a subset of $D$

for (a) I used $|A \times B|$ = $|A| * |B|$ and got $3*2 = 6$

is this the correct way to do this?

for (b) I assumed that the cardinality was infinite since it involved the set of natural numbers, am I correct in assuming this?

for part 2 (a) I assumed that it was true since $D$ contains the natural set so presumably the natural set is a subset of $D$, am I correct in assuming this?

Martin Sleziak
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jn025
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3 Answers3

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$\mathbb N$ is not a subset of $D$. Subsets of $D$ alike to $\mathbb N$ are, for example $$\{1\} \times \mathbb N, \{5\} \times \mathbb N$$ The difference is that elements from $D$ look like $(1, a)$ or $(5, a)$ with $a\in\mathbb N$ whereas elements from $\mathbb N$ are just natural numbers (no tuples).

As for the cardinalities, you are right; $$|A\times B| = 6, |A\times D| = |D| = |\mathbb N| = \aleph_0 \quad \text{("countable infinity")}$$

More generally spoken, there are subsets of $A\times B$ looking like $A$ or $B$, namely sets of the form $A\times \{b\}, \{a\} \times B$ with $a\in A, b\in B$, but $A,B$ are no subsets of $A\times B$.

AlexR
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For (a) and (b), you were right, but more specifically, the cardinality of $A \times D$ is $\aleph_0$, or countable infinity. (The same cardinality as $\Bbb N$.

For part 2 (a), you were wrong, however. $D$ does not contain $\Bbb N$, because $1 \neq (b,n)$ for any $b \in B, n \in \Bbb N$. Put more simply, $A \times B$ does not contain either $A$ or $B$ for any non-empty $A,B$.

Fargle
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|A x B| = |A| * |B| = 6 , but |A x D| = aleph_null i.e set A x D and set of nauturals are equipotent i.e there exists a bijection between A x D and N. note D is not a subset of N because elements of D are (1, 1) , (1,2) , (1,3),... , (5,1), (5,2),(5,3) ... i.e elements of D are in 2-tuple wheres elements of N are simply as usual numbers, so N is not subset of D

Akash Patalwanshi
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  • What does this add to the other answers give 4 hours ago? Also refer to this help page on MathJax to improve readability of your answers. – AlexR Aug 14 '14 at 12:42
  • Yes, my answer added not more things but i have explained in my answer that if there exists bijection from set to set of naturals then that set is equipotent to set of naturals and have the cardinality aleph null and thank you so much for providing me that page, i am a new here. – Akash Patalwanshi Aug 14 '14 at 12:49