Problem: Suppose $A \subset M$ is connected where $M$ is a metric space. Show that $A, \emptyset $ are the only subsets of $A$ that are clopen relative to $A$.

My thoughts:

The empty set is always open and closed. Since $M$ is closed and open, and $A=A \cap M$, then $A$ is open relative to $A$ (and similarly closed).

Now supposing there was another clopen set $U \neq \emptyset, A$, Then I am imagining I need to use $U$ and $U^c$ to show that there is a separation of $A$. My definition of a separation is that $A = U \cup U^C$ where $U \cap closure(U^c) = \emptyset$ and $closure(U) \cap U^c = \emptyset$ which would contradict that $A$ is connected.

If $U$ is clopen relative to $A$, then there exists $V$ open in $M$ with $U=V \cap A$, and $W$ closed with $U=W\cap A$, how can I show $U \cap closure(U^c) = \emptyset$ and $closure(U) \cap U^c = \emptyset$ ?

Edit: Worth noting this problem has been asked many times but not with relative topology :)

  • $\begingroup$ You are on the right track. You have a non-empty $U\subset A$ and by assumption, $U$ is clopen. That means $U^c$ is also clopen. Therefore, $A=U\cup U^c$. Since both $U$ and $U^c$ are open , $U$ is non-empty and $A$ is connected, $U^c$ must be empty which means $U=A$. $\endgroup$
    – John Douma
    Commented May 27, 2018 at 18:43
  • $\begingroup$ $U$ and $U^c$ are open in $M$ or the relative topology? It seems to me they are not necessarily open in $M$. $\endgroup$ Commented May 27, 2018 at 18:45
  • $\begingroup$ $M$ doesn't have to be connected. We are only interested in sets that are clopen relative to $A$. $\endgroup$
    – John Douma
    Commented May 27, 2018 at 18:47
  • $\begingroup$ Yes but what I am saying is that $U, U^c$ are only clopen relative to A, not shown to be open in $M$, which means your use of $A$'s connectedness won't work I believe. A is connected in $M$ after all. Perhaps I am confused. $\endgroup$ Commented May 27, 2018 at 18:54
  • $\begingroup$ In reading the entry on relative topology mathworld.wolfram.com/RelativeTopology.html I see that a set can be open in the relative topology and not the overall topology. $\endgroup$ Commented May 27, 2018 at 18:55

2 Answers 2


Note: I'm going to use a bar over a set to represent closure. (Thus the closure of $U$ is $\overline U$.) Also, complements will always be relative to $A$, but open/closedness relative to $M$ (matching the usage in your question).

$U^c = V^c \cap A$, so $U^c \subset V^c$, thus $\overline {U^c} \subset \overline {V^c} =V^c$ (since $V^c$ is closed). Thus $\overline {U^c} \cap V = \varnothing$, so $\overline {U^c} \cap U = \overline {U^c} \cap V \cap A = \varnothing$.

Similarly, since $U \subset W$, $\overline {U} \subset \overline {W} =W$, so $\overline U \cap W^c = \varnothing$. $U^c = W^c \cap A$, thus $\overline {U} \cap U^c = \overline {U} \cap W^c \cap A = \varnothing$.

  • $\begingroup$ Great ! thanks so much. $\endgroup$ Commented May 28, 2018 at 4:45

You don't need to use closure, and as you are working relative to $A$, you can forget about $M$ and work in $A$.

Suppose you have a clopen $U\subset A$. Then $U^c$ (relative to $A$) is also clopen. We now have $A=U\cup U^c$, with $U\cap U^c=\varnothing$. As both $U$ and $U^c$ are open and $A$ is connected, this means $U=\varnothing$ or $U^c=\varnothing$. Thus $U=\varnothing$ or $U=A$.

  • $\begingroup$ $U$ and $U^c$ are only open relative to $A$, not necessarily open in $M$, the definition of connected in $M$ is that there are not two open sets that separate $A$ in $M$ I believe. $\endgroup$ Commented May 27, 2018 at 18:47
  • $\begingroup$ On Wikipedia you can find that "A subset of a topological space $X$ is a connected set if it is a connected space when viewed as a subspace of $X$." I'm pretty sure that means $U$ and $U^c$ have to be open relative to $A$. (I know Wikipedia isn't the best source, but this definition is the same as in my lecture notes, which aren't in english.) $\endgroup$
    – TitiMcMath
    Commented May 27, 2018 at 20:26
  • $\begingroup$ Interesting ok. So it seems I need to show some of my definitions are equivalent to other definitions. If no one else answers this today I will accept later on. Thx. $\endgroup$ Commented May 27, 2018 at 20:27

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