"MATEMĀTIKA 10. KLASEI"
Par divu kopu A un B šķēlumu sauc tādu kopu, kura satur tikai tos elementus, kas pieder gan kopai A, gan kopai B. To pieraksta: $A\cap B$
Par kopu A un B apvienojumu sauc kopu, kas sastāv no visiem kopas A un visiem kopas B elementiem. To pieraksta: $A\cup B$. Kopu apvienojumā kopīgos elementus raksta tikai vienu reizi.

1. Ja $A=\left\{1;\phantom{\rule{0.147em}{0ex}}2;\phantom{\rule{0.147em}{0ex}}3;\phantom{\rule{0.147em}{0ex}}4;\phantom{\rule{0.147em}{0ex}}5\right\}$ un $B=\left\{3;\phantom{\rule{0.147em}{0ex}}4;\phantom{\rule{0.147em}{0ex}}5;\phantom{\rule{0.147em}{0ex}}6;\phantom{\rule{0.147em}{0ex}}7\right\}$, tad $A\cap B=\left\{3;\phantom{\rule{0.147em}{0ex}}4;\phantom{\rule{0.147em}{0ex}}5\right\}$, jo šie skaitļi ir kopīgi abām kopām, bet $A\cup B=\left\{1;\phantom{\rule{0.147em}{0ex}}2;\phantom{\rule{0.147em}{0ex}}3;\phantom{\rule{0.147em}{0ex}}4;\phantom{\rule{0.147em}{0ex}}5;\phantom{\rule{0.147em}{0ex}}6;\phantom{\rule{0.147em}{0ex}}7\right\}$.

2. $A=\left(-\mathrm{\infty };\phantom{\rule{0.147em}{0ex}}-2\right)$ un $B=\left(4;\phantom{\rule{0.147em}{0ex}}+\mathrm{\infty }\right)$, tad $A\cup B=\left(-\mathrm{\infty };\phantom{\rule{0.147em}{0ex}}-2\right)\cup \left(4;\phantom{\rule{0.147em}{0ex}}+\mathrm{\infty }\right)$, bet $A\cap B=\varnothing$.

3. Ja $A=\left(-\mathrm{\infty };\phantom{\rule{0.147em}{0ex}}-2\right)$, $B=\left(-4;\phantom{\rule{0.147em}{0ex}}0\right)$ un $C=\left(4;\phantom{\rule{0.147em}{0ex}}+\mathrm{\infty }\right)$, tad $A\cap B=\left(-4;\phantom{\rule{0.147em}{0ex}}-2\right)$, bet $A\cap B\cap C=\varnothing$.

Par kopu A un B starpību sauc kopu, kas sastāv tikai no tiem elementiem, kas pieder kopai A, bet nepieder kopai B. To pieraksta: $A\B$
Piemērs:
Ja $A=\left\{1;\phantom{\rule{0.147em}{0ex}}2;\phantom{\rule{0.147em}{0ex}}3;\phantom{\rule{0.147em}{0ex}}4;\phantom{\rule{0.147em}{0ex}}5\right\}$ un $B=\left\{4;\phantom{\rule{0.147em}{0ex}}5;\phantom{\rule{0.147em}{0ex}}6\right\}$, tad $A\B=\left\{1;\phantom{\rule{0.147em}{0ex}}2;\phantom{\rule{0.147em}{0ex}}3\right\}$, (no kopas $$A$$ elementiem izņēma visus tos, kas bija kopīgi ar kopu $$B$$).
Savukārt $B\A=\left\{6\right\}$.

$A\B$ nav tas pats, kas $B\A$. Salīdzini!

$B\A$