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Let 
    A = { x | x is an odd integer and 0 < x < 30 }  
    Which of the following are true?
  a. { 0,1,2,3 } ⊆ A
  b. { 5,3,1 } ⊆ A
  c.  ∅ ⊆ A
  d. A ⊆ { 5,3,1 }

Teacher said answers are : a. False , b. True , c. True , d. False

But for example in b, how can { 5,3,1 } be improper subset of A, if A has only 1 element?

And why is d False if it is possible that A might be either 5, 3 or 1, since it is can be an odd integer from 0 to 30?

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    $\begingroup$ $A=\{1,3,5,7,\cdots,27,29\}$ $\endgroup$ – Frudrururu Oct 25 '15 at 3:02
  • $\begingroup$ You should review set-builder notation. Not sure why this has down votes either. The asker has asked a question and shown their work. $\endgroup$ – Rocket Man Oct 25 '15 at 3:03
  • $\begingroup$ Hmm, I see, so set A consists of many elements, not 1. That explains a lot $\endgroup$ – Para Oct 25 '15 at 3:12
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The big issue, here, seems to be a misunderstanding of set-builder notation. When you see $$A=\{x\mid x\text{ is an odd integer and }0<x<30\},$$ you should interpret this to say: "$A$ is the set of all $x$ such that $x$ is an odd integer and $0<x<30.$" Put more plainly, $A$ is the set of all odd integers strictly between $1$ and $30,$ meaning that $A$ is the set of all odd integers from $1$ to $29$ (inclusive). Hopefully, having an appropriate conception of what $A$ is, in the first place, will make this problem more sensible.

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a) 0 not in set, and 2 not odd b) 5,3,1 are all odd and are all within 0 and 30. Also, it doesn't say anywhere that your set A is of only one element. A is all x, such that x follows the set of rules outlined. c) empty set is always a subset d) A is not possibly a subset of 5,3,1, since 7, 9, ... , 29 are not part of 5, 3 or 1

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