Definition of Open Manifolds in Jeffrey Lee's differential geometry book Definiton.
Manifold is an open manifold if it satisfies following to properties.

*

*It has empty boundary


*No component is compact.
For example, the interior int(M) of a connected manifold M with nonempty boundary is never compact and is an open manifold in the above sense if every component of M contains part of the boundary.
Questions

*

*M is assumed to be connected thus M has only one component i.e. itself. Thus every component of M contains part of the boundary.

I don't understand why the author stated "if every component of M contains part of the boundary."


*int(M) is certainly not compact and without boundary.
How can I prove that every component of int(M) is not compact?
Is int(M) connected?

 A: I think that the word "connected" in the "For example, ..." was not supposed to be there; the rest of the paragraph suggests that to me. Once you believe that, the rest makes good sense. 
Consider $M$ is the disjoint union of two line segments in the plane; the interior of $M$ is then the disjoin union of two open line segments in the plane, and the claim (that the interior of such a thing is an open manifold) is illustrated. 
To see why the author included the condition that each component contain some of the boundary, consider the case where $M$ is the union of a circle and a (closed) line-segment in the plane. The interior of $M$ (the circle union the open line segment) is not an open manifold, because the circle component is compact. And that's because the circle didn't contain any of the boundary of $M$. 
That's not to say that containing part of the boundary is a necessary condition -- it's merely a sufficient one. It's non-necessity is shown by the following:
Suppose we take as $M$ the union of that same closed interval and the whole $x$-axis. The interior of the union is in fact an open manifold, even though the $x$-axis part doesn't contain any boundary. And that's OK, because the author's claim is that if each component meets the boundary, then the interior is open, but not "only if". (I found the statement misleading as I read it, which is why I've included this apparently irrelevant example.)
