Application of connectedness

Question

I am reading connectedness in topology. Wondering where we can use this ? I have searched on internet but found only its role in motion planning robotics, population modeling.

Answer 1

I don't know much about topology, but I can give you some basic examples in which connectedness can give us useful information within mathematics.

Proposition. There are no homeomorphisms $f: [0,1) \rightarrow \mathbb{R}$. That is, there are no bijective functions $f: [0,1) \rightarrow \mathbb{R}$ such that $f$ and $f^{-1}$ are continuous.

Proof. Suppose there exists such a function $f$ and let $a = f(0)$. Since $f$ is bijective,

$$ f(0,1) = f([0,1) \setminus \{0\}) = f([0,1)) \setminus \{f(0)\} = \mathbb{R}\setminus \{a\} $$

However, since $f$ is continuous and $(0,1)$ connected, $f(0,1)$ should be connected, and so we have reached a contradiction.

Here, what may at first have seemed a very open ended problem, had a very succinct resolution via the fact that continuous functions preserve connectedness. It can even be surprising at first, if one is not familiar with these type of arguments, because it is well known that there are homeomorphisms between $(0,1)$ and $\mathbb{R}$.

Another fact that I find quite interesting is the following,

Proposition. Let $(X,d)$ be a metric space such that for any open ball, its points are path-connected to its center. Then, every connected open set of $X$ is path-connected.

Proof. Let $U \subseteq X$ be a non-empty, connected open set and let $x \in U$. Now, consider:

$$ A = \{y \in U : \text{there exists a path } \gamma \text{ from } x \text{ to } y \} $$

That is, formalized,

$$ A = \{y \in U : \exists\gamma \in C([0,1],X) \text{ s.t. } \gamma(0) = x, \gamma(1) = y\} $$

Now, $A$ is open in $U$: if $z \in A$ then there exists $r > 0$ such that $B_r(z) \subseteq U$ because $U$ is open. Each point of the ball is connected to $z$ by hyposthesis, and since $z \in A$, it is connected to $x$; so every point in $B_r(z)$ is connected to $x$. That is, $B_r(z) \subseteq A$.

Moreover, $A$ is also closed in $U$, if $z \in U \setminus A$, since $U$ is open there exists $r'$ such that $B_{r'}(z) \subseteq U$. Since $z$ is connected to every point in the ball, they cannot be connected to $x$ because that would make $z$ be connected to $x$, but $z \not \in A$. Thus, $B_{r'}(z) \subseteq U \setminus A$.

Since $A$ is closed and open in $U$ and $U$ is connected, either $A$ is empty or $A$ = $U$. However, $x \in A$, so necessarily $U$ equals $A$. Finally, if $z,z' \in U$, they are both connected to $x$, so joining paths we have that they are path-connected to each other, completing the proof.

In particular, in finite dimensional normed vector spaces over $\mathbb{R}$ or $\mathbb{C}$ open balls are convex, and so open connected sets are actually path-connected, which is a far more intuitive notion.

Finally, this next result proves that connected sets in metric spaces must be either trivial or at least as 'big' as the real numbers, cardinality-wise.

Proposition. Let $(X,d)$ be a connected metric space. Then either $\#X = 1$ or $\#X \geq \mathfrak{c}$.

We will need an auxiliary result first,

Lemma. Let $(X,d)$ be a connected metric space, $f: X \rightarrow \mathbb{R}$ a continuous function and $a \leq b \in f(X)$. Now, if $c \in [a,b]$, there exists $x \in X$ such that $f(x) = c$.

Proof. By the contrapositive, let's assume that there exits $c \in [a,b]$ such that $f(x) \neq c \quad (\forall x \in X)$ and conclude that $X$ is not connected. If it were the case, then we have that

$$ X = f^{-1}(-\infty, c) \cup f^{-1}(c,+\infty) $$

simply because for each point $x$, $f(x) \in \mathbb{R} \setminus \{c\}$. By continuity of $f$, each preimage set is open, and by hypothesis, they are both non empty because $a < c$ and $b > c$ are in the image of $f$. We have therefore written $X$ as a union of open, non empty, disjoint sets, which proves that $X$ is disconnected.

Now let's prove the actual proposition,

Proof. Let $x \in X$ and consider the continuous function $g := d(x, -) : X \rightarrow \mathbb{R}$. If $g$ is constant, then

$$ d(x,y) = d(x,x) = 0 $$

for all $y \in X$ and so $X = \{x\}$. Otherwise, there exist $a < b$ in the image of $g$. By the lemma, then, we have a function $\Gamma : [a,b] \rightarrow X$ that assigns $c \in [a,b]$ to a point $\Gamma(c) \in X$ such that $g(\Gamma(c)) = c$. It will suffice to show that $\Gamma$ is injective, in which case we will have that $\mathfrak{c} = \#[a,b] \leq \#X$.

In effect, if $\Gamma(c) = \Gamma(d)$, then

$$ c = g(\Gamma(c)) = g(\Gamma(d)) = d $$

or in other words, $g$ is a left inverse for $\Gamma$.