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Homede Morgan's laws for sets (proof)

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# de Morgan’s laws for sets (proof)

Let $X$ be a set with subsets $A_{i}\subset X$ for $i\in I$, where $I$ is an arbitrary index-set. In other words, $I$ can be finite, countable, or uncountable. We first show that

$\displaystyle\displaystyle\big(\cup_{{i\in I}}A_{i}\big)^{{\prime}}$ | $\displaystyle=$ | $\displaystyle\cap_{{i\in I}}A_{i}^{{\prime}},$ |

where $A^{{\prime}}$ denotes the complement of $A$.

Let us define $S=\big(\cup_{{i\in I}}A_{i}\big)^{{\prime}}$ and $T=\cap_{{i\in I}}A_{i}^{{\prime}}$. To establish the equality $S=T$, we shall use a standard argument for proving equalities in set theory. Namely, we show that $S\subset T$ and $T\subset S$. For the first claim, suppose $x$ is an element in $S$. Then $x\notin\cup_{{i\in I}}A_{i}$, so $x\notin A_{i}$ for any $i\in I$. Hence $x\in A_{i}^{{\prime}}$ for all $i\in I$, and $x\in\cap_{{i\in I}}A_{i}^{{\prime}}=T$. Conversely, suppose $x$ is an element in $T=\cap_{{i\in I}}A_{i}^{{\prime}}$. Then $x\in A_{i}^{{\prime}}$ for all $i\in I$. Hence $x\notin A_{i}$ for any $i\in I$, so $x\notin\cup_{{i\in I}}A_{i}$, and $x\in S$.

The second claim,

$\displaystyle\big(\cap_{{i\in I}}A_{i}\big)^{{\prime}}$ | $\displaystyle=$ | $\displaystyle\cup_{{i\in I}}A_{i}^{{\prime}},$ |

follows by applying the first claim to the sets $A_{i}^{{\prime}}$.

## Mathematics Subject Classification

03E30*no label found*

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