# In a metric space X, if A is connected, is its interior connected?

I believe that int(A) is also connected. I tried to use an argument by contradiction, that is to say I supposed that int(A) is not conneted, but without success

• umm It's interior is the empty set – DIEGO R. Jan 9 '17 at 5:47
• @MyGlasses Sets that has empty interior would not be any good example for this as the empty set is connected (as any set with no more than one element trivially is). Such sets could only be used as example of the opposite: non-connected sets with connected interior (for example $\{0,1\}$ which is non-connected, but has connected interior). – skyking Jan 9 '17 at 7:21
• Consider the union of first and third quadrants on the plane along with the axes. – Henricus V. Jan 9 '17 at 7:34
• Note that it is a fairly reasonable convention that the empty set is not connected (it has zero connected components, not one). Of course this is not an ideal example, since it immediately raises the question of whether requiring the interior to be nonempty changes the answer. (No, as the comment above and the answer below show.) – Pete L. Clark Jan 9 '17 at 14:04
• For $n \ge 2$, you can take any two connected sets in $\mathbb R^n$ with non-empty interiors, whose union is not connected, and join them by a straight line. – TonyK Jan 9 '17 at 20:31

## 2 Answers

How about this? Two closed disks tangent to each other at a point. • I got it!. Int (A) is not necessarily connected – DIEGO R. Jan 9 '17 at 5:45
• This is one of those situations where a visual demonstration really drives the point home. Would +1 a second time if I could. – Robin Saunders Jan 9 '17 at 23:10

Consider $X=\mathbb R^2$ and $$A=([-2,0]\times[-2,0])\cup([0,2]\times[0,2])$$which is connected, while $\text{int}(A)$ is not connected.

To see this consider the continuous function $f:\mathbb R^2\to\mathbb R$ is defined by $f(x,y)=x+y$. Let $U=f^{-1}(0,+\infty)$ which is open in $\mathbb R^2$ and so $U\cap\text{int}(A)$ is open in $\text{int}(A)$. Also, since $(0,0)\notin\text{int}(A)$, so for all $(x,y)\in\text{int}(A)$, $f(x,y)\neq0$ and $U\cap\text{int}(A)=f^{-1}[0,+\infty)\cap\text{int}(A)$ is closed in $\text{int}(A)$. Furthermore, $(1,1)=f^{-1}(2)\in U\cap\text{int}(A)$ shows that $U\cap\text{int}(A)\neq\emptyset$ while $(-1,-1)\in\text{int}(A)$ and $(-1,-1)\notin U$ shows that $U\cap\text{int}(A)\neq\text{int}(A)$.