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Consider the set $A = \{x \in \mathbb{Z{\ge_0}}$ and $x \in {-2}\}, B = \{f(x)\}$.

Is $f: A \to B$ still a bijection even though $f(-2)$ is undefined?

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  • $\begingroup$ Try the command \text{ and } to insert a text "and" in math mode. Or use \land (logical and) in math mode for a $\land$. $\endgroup$ – Theo Bendit Feb 11 at 15:41
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    $\begingroup$ I'm still confused as to the actual question. Is $A$ a set? A function? If the latter, what is its domain? And what does $x \in -2$ mean exactly? If you're looking for curly braces to help define a set, try \{ and \} in math mode. $\endgroup$ – Theo Bendit Feb 11 at 15:44
  • $\begingroup$ Your question is not at all clear. What are the two sets that you claim are in bijection? What is the proposed bijection between them? $\endgroup$ – lulu Feb 11 at 15:46
  • $\begingroup$ @TheoBendit see edit $\endgroup$ – Jossie Calderon Feb 11 at 15:57
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    $\begingroup$ Voting to close the question as it is not clear what you are asking. If you can, please edit for clarity. $\endgroup$ – lulu Feb 11 at 16:32
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A function $f: A \rightarrow B$ is a triple ($f, A, B$) where $f \subseteq A \times B$ satisfying certain properties. So what you should really be asking is, is $f$ a bijection between A and B? If $f(-2)$ is not defined, this means that $-2 \notin A$, so it doesn't have any consequence on whether your function is a bijection between A and B or not.

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