# Two tangent closed discs connected

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 $X$.

• What do you want to prove exactly? Commented Jun 19, 2013 at 20:46
• @PatrickDaSilva That two tangent closed balls are connected.
– Pedro
Commented Jun 19, 2013 at 20:46
• As usual, it's easier to prove that $X$ is path-connected. Commented Jun 19, 2013 at 20:47
• @Chris : You answered the question then. Commented Jun 19, 2013 at 20:48
• And if you think using pathconnectedness is cheating, perhaps you know the following fact, which is fairly easy to prove and may be used if you believe that the closed discs themselves are connected: If $\{A_\alpha \mid \alpha \in I\}$ is a family of connected subsets of a topological (or metric) space $M$, and $A_\alpha \cap A_{\alpha'} \not= \emptyset$ for all $\alpha,\alpha' \in I$, then $\bigcup_{\alpha \in I} A_\alpha$ is connected. Commented Jun 19, 2013 at 20:54

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.
Particularly, if we have two connected subspaces $X, Y\subset Z$ such that $X$ and $Y$ aren't disjoint, then $X\cup Y$ is connected (this obviously applies to your case since $X=Y=D^1$ are tangent). Say we have a separation of $Z$, i.e. disjoint open sets $U, V$ whose union contains $Z$. Then since $X$ is connected, either $X\subset U$ or $X\subset V$ (can you see why?) and similarly for $Y$. Moreover, since $X$ and $Y$ aren't disjoint, they need to be contained within the same set, either $U$ or $V$. But this means that $Z$ is contained in either $U$ or $V$, so $Z$ is connected. This is all in Munkres, Introduction to Topology in the chapter on connectivity.
This also works if we have an arbitrary collection of connected subspaces $\{X_i\}_{i\in I}$ which are pairwise non-disjoint.