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Hello everyone assume that we set $A$ = {{1 ,2} , {2 ,3} ,{4 ,3}}

so what is $\cup_{B \in A} (B)$?

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  • $\begingroup$ You may be confused by the notation. $\cup_{B\in A} (B)$ means you have to take the union of the elements of $A$, in this case $\cup_{B\in A} (B) = \{1,2\} \cup \{2,3\} \cup \{4,3\}$ $\endgroup$
    – jjagmath
    Aug 1 '20 at 10:52
  • $\begingroup$ So what is the difference from $\cup{A}$? $\endgroup$
    – xx01
    Aug 1 '20 at 10:54
  • $\begingroup$ $\cup A$ is a more compact notation for the same thing. It makes sense especially in axiomatic set theory, where everything is a set (including the elements of a set). $\endgroup$
    – Aurelio
    Aug 1 '20 at 11:00
  • $\begingroup$ Thank you very much!!! $\endgroup$
    – xx01
    Aug 1 '20 at 11:00
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$\{1,2\}\cup \{2,3\} \cup \{4,3\}$..

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  • $\begingroup$ So what is the difference from $\cup{A}$? $\endgroup$
    – xx01
    Aug 1 '20 at 10:54
  • $\begingroup$ Depends on what you want it to stand for. The subscripts are required to avoid confusion: it may very well be $\cup_{A \in \mathcal A} A$. In the absence of a subscript, some authors may say that union is taken over elements of $A$ as in above. $\endgroup$
    – Coward
    Aug 1 '20 at 10:58
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Here The elements of $A$ are $\{1,2\}, \{2,3\}, \{4,3\}$. So the union of elements of $A$ is:

$$\cup_{B\in A} B = \{1,2\} \cup \{2,3\} \cup \{4,3\}.$$

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I surmise that this was a trick question. The intention is for the student to realize that the union of all elements of a specific set is the specific set itself, without any regard to what the elements actually are.

Addendum

Subsequent comments by Aurelio and Coward have suggested that I am mistaken. I may well be mistaken, but I'm just not seeing it yet.

Coward: In response to your comment, could you please provide a counter example?

Aurelio: In response to your counter example, perhaps I'm missing something here. In your counter example, it does seem to me that $\bigcup_{b \in A} \;=\; A.$ Please explain if you think that I am mistaken.

Addendum-2 In response to John Hughes comment below:
Well this is an eye-opener. His explanation, which never occurred to me, does seem sensible.
I now have to say that (apparently) my knowledge of set theory (re interpretation of notation) is deficient, so my answer (above) may well be wrong.

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    $\begingroup$ Are you saying that $\cup A = A$? That's not true. $\endgroup$
    – Coward
    Aug 1 '20 at 11:04
  • $\begingroup$ I am not convinced by this phrasing. I suppose you mean that the union of a cover of a set is the set itself. For example, for $A=\{\{0\},\{1,2\}\}$, $\cup A\neq A$. $\endgroup$
    – Aurelio
    Aug 1 '20 at 11:04
  • $\begingroup$ If I am not mistaken, what you want to say is that a set is the union of its singleton-subsets, which is almost a way of stating the axiom of extensionality. $\endgroup$
    – Aurelio
    Aug 1 '20 at 11:12
  • $\begingroup$ @Coward Please see my Addendum. $\endgroup$ Aug 1 '20 at 11:13
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    $\begingroup$ Your answer would be correct if the question had been $\cup_{b \subset A} b$ rather than $\cup_{b \in A} b$. The notation written says to take the union of the elements of the set $A$, not its one-element subsets. (My own replacement says to take the union of all subsets, not just one-element ones, by the way.) $\endgroup$ Aug 1 '20 at 11:18

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