### Theory:

Disjoint set:
Two sets $$A$$ and $$B$$ are said to be disjoint if they do not have any element in common for two sets $$A$$ and $$B$$.

Symbolically we write $$A\cap B=\varnothing$$ (where $$\varnothing$$ is empty set).

$\begin{array}{l}A\phantom{\rule{0.147em}{0ex}}= \left\{2,\phantom{\rule{0.147em}{0ex}}4,\phantom{\rule{0.147em}{0ex}}6,\phantom{\rule{0.147em}{0ex}}8\right\},\phantom{\rule{0.147em}{0ex}}B\phantom{\rule{0.147em}{0ex}}= \left\{10,\phantom{\rule{0.147em}{0ex}}12\right\}.\\ \\ A\cap B=\phantom{\rule{0.147em}{0ex}}\left\{2,\phantom{\rule{0.147em}{0ex}}4,\phantom{\rule{0.147em}{0ex}}6,\phantom{\rule{0.147em}{0ex}}8\right\}\cap \phantom{\rule{0.147em}{0ex}}\left\{10,\phantom{\rule{0.147em}{0ex}}12\right\}\end{array}$

Where no common elements are in the set $$A$$ and $$B$$.

$$A$$ and $$B$$ are disjoint.

So, $$A\cap B=\varnothing$$

Important!
$$A$$ and $$B$$ do not have any element in common and are disjoint sets.
Overlapping sets:
Two sets $$A$$ and $$B$$ are said to be overlapping if they contain at least one element in common from two sets $$A$$ and $$B$$.

Symbolically we write $$A\cap B\neq \varnothing$$ (where $$\varnothing$$ is empty set\).

$\begin{array}{l}A=\left\{\phantom{\rule{0.147em}{0ex}}2,\phantom{\rule{0.147em}{0ex}}4,\phantom{\rule{0.147em}{0ex}}6,\phantom{\rule{0.147em}{0ex}}8\right\},\phantom{\rule{0.147em}{0ex}}B\phantom{\rule{0.147em}{0ex}}= \left\{\phantom{\rule{0.147em}{0ex}}6,\phantom{\rule{0.147em}{0ex}}8,\phantom{\rule{0.147em}{0ex}}10,\phantom{\rule{0.147em}{0ex}}12\right\}.\\ \\ A\cap B=\phantom{\rule{0.147em}{0ex}}\left\{\phantom{\rule{0.147em}{0ex}}2,\phantom{\rule{0.147em}{0ex}}4,\phantom{\rule{0.147em}{0ex}}6,\phantom{\rule{0.147em}{0ex}}8\right\}\cap B\phantom{\rule{0.147em}{0ex}}= \left\{\phantom{\rule{0.147em}{0ex}}6,\phantom{\rule{0.147em}{0ex}}8,\phantom{\rule{0.147em}{0ex}}10,\phantom{\rule{0.147em}{0ex}}12\right\}\end{array}$

Where, at least one common element is in both the set $$A$$ and $$B$$.

Because $$A$$ and $$B$$ are overlapping.

So, $$A\cap B=\{6,\ 8\}\neq\varnothing$$

Therefore, $$A\cap B\neq\varnothing$$

In the Venn diagram below, the area in the blue colour represents $A\cap B\ne \varnothing$.