# more intuitive definition of connectedness

I know what does connectedness mean theorically but I'm facing some difficulties trying to understand its concept it's too abstract for me

can someone provide a trivial/simple example of a set that is connected and/or explain me the concept of connectedness like I'm five.

• Do you know the definition of disconnectedness? – Arpit Kansal Mar 18 '17 at 10:36
• Any interval is connected. Any cartesian product of intervals is connected. Any set defined in $\Bbb R^n$ by the points of a continuous curve (in a connected domain) is connected. – Masacroso Mar 18 '17 at 10:38
• @Arpit Kansal No I first read it as 'connectedness' sorry going to google it – rapidracim Mar 18 '17 at 10:43

$[0,1]$ is a good example of a connected space.

$[0,1] \cup [2,3]$ is a good example of a disconnnected space.

Intuitively, $[0,1] \cup [2,3]$ is disconnected because it has a "gap" in the middle. $[0,1]$ has no gap.

Perhaps the most intuitive definition of connectedness is the following: A topological space $X$ is connected iff there are no continuous maps $f : X \to \{ 0, 1 \}$ apart from the two obvious constant maps, $f(x) = 0$ and $f(x) = 1$. Said another way, $X$ is connected iff there are no continuous surjective maps $f : X \to \{ 0, 1 \}$.

You'll notice that, indeed, the only continuous maps $f: [0,1] \to \{ 0, 1\}$ are the maps $f(x) = 0$ and $f(x) = 1$. This follows from the Intermediate Value theorem.

However, there is a continuous map $f : [0,1] \cup [2,3] \to \{ 0, 1 \}$ given by $$f(x) = \begin{cases} 0 & x \in [0,1]; \\ 1 & x \in [2,3].\end{cases}$$ So $[0,1] \cup [2,3]$ is disconnected, by my definition. Intuitively, the "gap" in the set $[0,1] \cup [2,3]$ is precisely the feature that enables us to define a continuous map to $\{0, 1\}$ in this way!

But how does my definition relate to the usual definition of connectedness? The usual definition is: $X$ is connected iff it is impossible to write $X$ as a disjoint union of two non-empty open subsets.

Well, suppose $X$ is disconnected by my first definition. Then there exists a surjective map $f: X \to \{ 0, 1\}$. But then, $X$ is the disjoint union of the non-empty open sets $f^{-1}(0)$ and $f^{-1}(1)$.

Conversely, suppose $X$ is disconnected by your usual definition, i.e. $X$ can be written as a disjoint union of the non-empty open sets $U$ and $V$. Then the function $X \to \{ 0, 1\}$ sending $U$ to $0$ and $V$ to $1$ is a surjective, continuous function from $X$ to $\{ 0, 1\}$.

In our example of $X = [0,1] \cup [2,3]$, the two non-empty, disjoint open subsets are $U = [0,1]$ and $V = [2,3]$. (Note that $U$ and $V$ are open when considered as subsets of $X$, but are not open when considered as subsets of $\mathbb R$.) Our example of a surjective continuous function $f: X \to \{0, 1\}$ sends $U$ to $0$ and $V$ to $1$.