Let $X$ be a connected subset of a metric space $M$. Show that $X^0$ (the interior of $X$) is not necessarily connected.
So the example I'm thinking of is $X$ being two closed discs, tangent at a point. The interior is clearly not connected, since there exist two disjoint, non-empty open sets whose union is the interior (the two sets are exactly the two discs.)
But I don't know how I can prove that $X$ is connected. Using the definition of connectedness, I must prove either
(i) The only subsets of $X$ which are both open and closed are $\emptyset$ and $X$
or
(ii) If $A,B$ are disjoint open subsets of $X$ whose union is $X$, one of them contains $X$.
Best Answer
Connectedness can be a bit of an abstruse concept to work with. It's often easier to work with the stronger concept of path-connectedness (a space is path-connected if any two points can be joined by a continuous path in the space). Not every connected space is path-connected, but for those that are, this is generally the easiest way to prove connectedness.
In this case, for example, it's almost trivial to see that $X$ is path-connected: two points in the same disc can be joined by a straight-line path, while two points in different discs can be joined by a composite path formed of two line segments meeting at the tangency point.