What is an overlap?

Question

I want to ask what an overlap is.

My teacher said that for example $1$: Everything is an overlap hence it is not locally finite.

For example $2$, it doesnt overlap.

Please teach me these two examples.

Thanks.

Answer 1

There are two families of sets mentioned on that page.

• One is $\left\{\left(-\frac1n,\frac1n\right):n\in\Bbb Z^+\right\}$; it’s an open cover of $(-1,1)$, and it is not even point-finite, since $0$ belongs to all of the sets. The sets $U_p$ mentioned just below it do not seem to have anything to do with it, however, or with anything else on the page.

• The other is $\mathscr{F}=\{[n,n+1]:n\in\Bbb Z^+\}$; it’s a closed cover of $[1,\to)$ (or if you prefer, $[1,\infty)$), and it is locally finite. To see this, suppose that $x\ge 1$. If $x$ is not an integer, then there is a positive integer $n$ such that $n<x<n+1$; let $U_x=(n,n+1)$. Then $U_x$ is an open neighborhood of $x$, and the only member of $\mathscr{F}$ that intersects $U_x$ is $[n,n+1]$: if $m\in\Bbb Z^+$, and $m\ne n$, then $[m,m+1]\cap(n,n+1)=\varnothing$, because either $m<n$, in which case $m+1\le n$, or $n<m$, in which case $n+1\le m$. If $x$ is an integer, let $U_x=(x-1,x+1)$. Suppose that $n\in\Bbb Z^+$ and $U_x\cap[n,n+1]\ne\varnothing$; then either $x=n$, or $x=n+1$, since $x$ is the only integer in $U_x$. In either case, then, $U_x$ intersects at most two members of $\mathscr{F}$. Thus, every $x\ge 1$ has an open neighborhood that intersects only finitely many members of $\mathscr{F}$, which by definition means that $\mathscr{F}$ is locally finite.

Finally, it appears that your teacher means that sets $A$ and $B$ overlap if $A\cap B$ contains an open interval, i.e., is more than just one point. With this definition $(0,2]$ and $[1,3)$ overlap, because their intersection, $[1,2]$, contains the open interval $(1,2)$, but $[0,1]$ and $[1,2]$ do not overlap, because $\{1\}$, their intersection, does not contain an open interval.