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Can you help me understand what the exponent means here?

$$ (A\cap B) \cup D^C $$

A, B, C and D are all sets of a universal set U.

Thank you.

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Are you sure that $C$ is a set, and not a symbol for the complement? In general, $X^Y$ often means "the set of functions $Y \to X$", perhaps with some additional properties like continuity. That doesn't seem to be the case here. –  Dylan Moreland Sep 12 '11 at 19:14
    
I'm pretty sure it must be the complement. I can close this unless you would like to make an answer. –  Louis Sep 12 '11 at 19:23
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3 Answers

up vote 4 down vote accepted

Small lower-case $c$ as an exponent when a Universe of Discourse is understood generally represents complement relative to $U$. That seems to be almost certainly the meaning in what you write: $(A\cap B)\cup D^c$ would be the collection of all things that are in either in both $A$ and $B$, or else that are not in $D$, as $$D^c = \{x\in U \mid x\notin D\}.$$ If that's the case (a lower-case $c$), then $D^c$ is pronounced "complement of $D$" or "$D$-complement."

Note the difference between $$D^c\quad\text{and}\quad D^C.$$ If $C$ is a set, then $D^C$ is, as others have noted, the set of all functions with domain $C$ and image contained in $D$, $$D^C = \{f\colon C\to D\mid f\text{ is a function}\}.$$

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In the general set theoretic settings, $X^Y=\{f\colon Y\to X\mid f\ \text{is a function}\}$. That is, this is the set of all functions with domain $Y$ and range which is a subset of $X$.

In the introductory context, in which there is usually a universal set being involved (although these can be involved in an explicit way much further into mathematics, however I doubt you would have asked this question in that situation), it is common to denote with a lowercase $c$ the complement with respect to the universal set, that is $D^c=U\setminus D=\{x\in U\mid x\notin D\}$, where $U$ is the universal set.

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Generally, when we write $D^C$ for two sets $D,C$, what we mean is the set of functions $f:C \rightarrow D$. Consider for example $C$ being a two-element set, $\{ 1,2 \}$. Then the set of functions from $C$ to $D$ is just the set of ordered pairs $(d_1,d_2)$ of elements of $D$ -- and hence has cardinality $|D|^2$. In general, the cardinality of $D^C$ is $|D|^{|C|}$ for finite sets, and we use that to generalize cardinality exponentiation for infinite sets.

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Thank you for your answer, but as Dylan Moreland suggested, it's probably the complement (the course hasn't covered anything in your answer). –  Louis Sep 12 '11 at 19:24
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