# Question about definition of connectedness.

While reading some books, I often find the following argument:

Consider $A\subset X$, when $X$ is a topological space.

If $A$ is a disconnected set, then there exist some open sets, say $U,V \in \tau_X$ disjoint and non empty such that $A = U \cup V$.

But this often confuses me, because the part when it says "open sets of $\tau_X$" because, for example, $[0,1] \cup [5,6]$ is disconnected and it can't be written as union of two open sets -of X!- which are disjoint and non empty.

The definition in some books of disconnected set $A$ says that $A$ is disconnected if it is disconnected with the subspace topology.

But then, I don't understand why some authors take open sets of the big space, say $X$.

Can someone help me to clarify this?

• I'm guessing you're misreading something. Can you give an exact quote? – Eric Wofsey Aug 6 '17 at 3:41
• For example, in the book "Topics on Continua" of Sergio Macías, in page 52 it says "Suppose that $X\setminus A$ is not connected. Then there exist two nonempty disjoint open subsets $U$ and $V$ of $X$ such that $X\setminus A = U \cup V$ ..." – HeMan Aug 6 '17 at 3:44
• "...because, for example, $[0,1]∪[5,6]$ is disconnected and it can't be written as union of two open sets -of $X$!" - what is $X$ here? – user 170039 Aug 6 '17 at 3:45
• Presumably in that context $A$ is closed, so $X\setminus A$ is open and relatively open sets are the same as open sets of $X$. – Eric Wofsey Aug 6 '17 at 3:45
• In that particular case, $X= \mathbb{R}$ – HeMan Aug 6 '17 at 3:47

The correct definition of $A$ being a disconnected subset of $X$ is that $A$ is disconnected as a space in its own right, i.e. having the subspace topology.

If $A \subseteq X$ is a disconnected subset, we can translate this to a statement about open sets of $X$ if we like:

$A \subseteq X$ is disconnected iff there exist $U,V$, open sets in $X$, such that

• $U \cap A \neq \emptyset$
• $V \cap A \neq \emptyset$
• $A \subseteq U \cup V$
• $A \cap U \cap V = \emptyset$

This corresponds to $\{U \cap A$, $V \cap A\}$ being a disconnection of $A$ in its subspace topology.

Definition of Separation of a Topological Space. Let $(X,\tau_X)$ be a topological space and $Y\subseteq X$. Let the subspace topology on $Y$ be denoted as $\tau_Y$. A separation of $Y$ is a pair of disjoint non-empty open (open in $Y$, not $X$) subsets of $Y$ whose union is $Y$.

Definition of Disconnected Topological Space. A topological space $(X,\tau_X)$ is said to be disconnected if there exists a separation of $X$.

So when trying to prove that $[0,1]\cup [5,6]$ is disconnected we shouldn't be trying to write it as union of two sets open in $\mathbb{R}$, rather we should try to write it as an union of sets open in $[0,1]\cup [5,6]$ under the subspace topology induced by the topology on $\mathbb{R}$ under consideration.

Here is a way to clarify this..

Definition. A topological space $(X,\tau)$ is disconnected if there exist $U,V\in\tau\setminus\{\varnothing\}$ such that $U\cap V=\varnothing$ and $U\cup V = X$. Otherwise, $(X,\tau)$ is connected.

We say that a subset $Y\subseteq X$ is connected if it is connected as a topological subspace.

Proposition. Let $(X,\tau)$ be a topological space and $Y\subseteq X$. We have that $Y$ is disconnected if and only if there exist $U,V\in \tau\setminus\{\varnothing\}$ such that

(1) $U\cap Y\neq\varnothing$;

(2) $V\cap Y\neq\varnothing$;

(3) $U\cap V\cap Y=\varnothing$; and

(4) $Y\subseteq U\cup V$.

Proof. (ONLY IF) Assume $Y$ is disconnected, then there exist $U_Y, V_Y\in\tau_Y\setminus\{\varnothing\}$ such that $U_Y \cap V_Y = \varnothing$ and $U_Y\cup V_Y = Y$. Thus there exist $U,V\in\tau\setminus\{\varnothing\}$ such that $U_Y=U\cap Y$ and $V_Y=V\cap Y$. Such $U$ and $V$ satisfy conditions (1)-(4).

(IF) Assume there exist such $U$ and $V$ in $\tau$ satisfying (1)-(4). Let $U_Y=U\cap Y$ and $V_Y=V\cap Y$, both in $\tau_Y\setminus\{\varnothing\}$ (conditions (1) and (2) tell us that both $U_Y$ and $V_Y$ are nonempty). Condition (3) tell us that $U_Y\cap V_Y= \varnothing$ and, finally, condition (4) says that $U_Y\cup V_Y=Y$. Therefore, $Y$ is disconnected. $\square$