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I got the first few, but im not sure about these:

  1. {2,4}⊂A×A.

  2. {∅} ∈ P(A).

  3. (1,1)∈A×A.

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closed as off-topic by Eevee Trainer, Henning Makholm, Jyrki Lahtonen, Jean-Claude Arbaut, Cesareo Apr 22 at 9:36

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  • 3
    $\begingroup$ Please tell us your thoughts so we can help you better. $\endgroup$ – rogerl Apr 22 at 0:35
  • $\begingroup$ I think 7 and 8 is true. Not sure abt 6. $\endgroup$ – ionics Apr 22 at 0:45
  • $\begingroup$ It is a set.... $\endgroup$ – ionics Apr 22 at 0:46
  • $\begingroup$ Alright, that solves it. Thanks! $\endgroup$ – ionics Apr 22 at 0:50
  • $\begingroup$ @ionics Actually, my mistake. $A\times A = \{ (a, b) | a, b\in A\}$. But the point still stands, $A\times A$ is a set of ordered pairs, so a set of numbers cannot be a subset of it $\endgroup$ – Andrew Li Apr 22 at 0:55
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$6)F, 7)F, 8)T$.

$6)$: the elements of $A×A$ are ordered pairs, of the form $(x,y)$, where $x,y\in A$.

$7)$: $\emptyset\in P(A)$, for any set $A$, but $\{\emptyset\}\neq\emptyset$, and $\{\emptyset\}\not\in P(A)$ for this $A$.

$8)$: see $6)$.

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  • $\begingroup$ the element (2,4) belongs to $A$x$A$ but the set {2,4} does not belong to $A$x$A$ is this what you had in mind when you say 6 is false? Thanks. $\endgroup$ – NoChance Apr 22 at 4:45
  • 1
    $\begingroup$ Yeah. That's right. $\{2,4\}$ is neither an element nor a subset of $A×A$. We need ordered pairs. $\endgroup$ – Chris Custer Apr 22 at 4:49
  • $\begingroup$ I appreciate your clarification. Thanks. $\endgroup$ – NoChance Apr 22 at 4:49
  • $\begingroup$ Sure thing. @NoChance $\endgroup$ – Chris Custer Apr 22 at 4:52

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