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# Path-connectedness

Question: Can I travel anywhere I want in $X?$ If Yes, then $X$ is can be called "path-connected".

By travel from $A$ to $B$ we mean a continuous function $q: [0,1] → X$ such that $q(0)=A, q(1)=B$.

**Theorem.** Suppose function $f: X → Y$ is continuous. Then, if $X$ path-connected then so is its image, $f(X)$, under $f$.

*Proof.* The proof is very simple. For any two points in $f(X)$ you need to be able to travel from one to the other. But in $f(X)$ every point comes from $X$ under $f$. So those two points have to be $f(A), f(B)$ for some $A, B$ in $X$. Now since $X$ is connected, there is a continuous function $q: [0,1] → X$ such that $q(0)=A, q(1)=B$. Then $fq: [0,1] → f(X)$ is continuous and $fq(0)=f(A), fq(1)=f(B)$. $\blacksquare$

The construction generates an equivalence relation on $X$. The equivalence classes are called *connected components*.

It is easy to show that connectedness is a topological invariant. And so is the number of components, see Betti numbers.

Recall, how the fact that the domain of the function is made of two pieces affects its continuity:

- $D(f) = ( -\infty, 0 ] {\cup} [ 1, \infty )$, i.e. "two pieces" and $f$ is continuous!
- Even if undefined at the end points, it's still continuous.
- Even if these end points are the same point, still two pieces and still continuous. Here $1 \not\in D(f) = ( -\infty, 1 ) {\cup} ( 1, \infty )$
- This one is not continuous! $D(f) = ( -\infty, 1 ] {\cup} ( 1, \infty ) = {\bf R}$, one piece!

Let's prove 4. In the definition of continuity we let $$a = 1, x = 1 + \epsilon, \epsilon > 0,$$ then $$| f(x) - f(a) | = | 1 - 0 | = 1. $$ Done.

**Exercise.** Finish...

What is the point? Continuity of a function depends on the topology of its domain, specifically, its connectedness.

Suppose

Is there a continuous path from $P$ to $Q$?

**Definition.** A subset $X$ of ${\bf R}^n$ is called *path-connected* if for any $P, Q \in X$ there is a continuous function
$$p: [ a, b ] \rightarrow {\bf R}^n$$
such that
$$p(a) = P, p(b) = Q.$$

**Examples:**

**Theorem.** $X = {\bf R}^n$ is path-connected.

**Proof.** In order to prove the claim, we have to find the path from $P$ to $Q$ for any given $P, Q {\in} {\bf R}^n$. The idea is to use a straight line.
Define

(This is a convex combination of $P$ and $Q$.) Prove now that $p$ is continuous with

$\blacksquare$

**Exercise.** Let $X = {\bf B}^n$ denote a closed ball. Is it path-connected? Yes. Proof is same as above.

**Exercise.** Apply the same proof to show that any convex set is path-connected (think of a box, a square, a 3d cylinder, etc).

**Exercise.** The same proof cannot be applied for donuts or circles, since they have holes and aren't convex.

**Exercise.** Let
$$X = [ 0, 1 ] \times \{0 \} \cup \{0 \} \times [ 0, 1 ], $$
where

Let further $$P = ( 1, 0 ), Q = ( 0, 1 ).$$ Now let us find the path from $P$ to $Q$. Define piecewise

In order to prove continuity, we verify only at point $a = 1$. Finally,

$$p(0) = ( 1 - 0, 0 ) = ( 1, 0 ) = P,$$ $$p(2) = ( 0, 2 - 1 ) = ( 0, 1 ) = Q.$$

**Theorem.** If $A, B$ are path-connected and $A \cap B \neq \emptyset$, then $A \cup B$ is path-connected.

*Proof* The idea is to combine functions for $A$ and $B$. Given $P, Q \in A \cup B$, find the path. Pick $N \in A \cap B$. Suppose $P \in A, Q \in B$. From connectedness of $A$ and $B$ it follows that we can

Then combine $p$ and $q$ into a path from $P$ to $Q$. Define $r(t)$ piecewise with $r$ continuous, $r: [ ⋅ ] \rightarrow {\bf R}^n$. (**Exercise.**) $\blacksquare$

**Exercise.** Let $X$ be a circle, which is not convex. Prove that $X$ is connected. Give an explicit formula for the path.

**Theorem (Path-connectedness in ${\bf R}$).** In ${\bf R}$, the path-connected sets are:

- intervals (open, closed, half-open, finite, infinite),
- single points,
- the empty set,

and these only.

In order to prove the theorem, show

- These sets are connected, i.e., construct a continuous function on an interval (they are convex!).
- There are no other connected sets.

**Theorem.** The image of a path-connected set under a continuous function is path-connected, or:

and

then

Recall from calc1:

**Intermediate Value Theorem.** Let

Let further $$A \leq C \leq B.$$ Then there is $c \in [ a, b ]$ with $f(c) = C$.

Observe that if $A, B \in f( [ a, b ])$, then $C \in f( [ a, b ])$. What does it tell about the topology of $f( [ a, b ])$?

**Rephrased IVT**. $f( [ a, b ])$ is path-connected (moreover it's an interval).

Then more general is this.

**Theorem.** Let
$$f: {\bf R}^n \rightarrow {\bf R}^m$$
and
$$P \subset D(f) \subset {\bf R}^n$$
be path-connected. Then $f(P)$ is path-connected.

**Proof.** Idea: re-state the problem of finding the necessary path in $f(P)$ as a problem of finding a certain path in $P$.

Recall the *definition*: A set $Q$ is path-connected if for any $A, B \in Q$ there is a continuous function $p: [ a, b ] \rightarrow Q$ such that

Set $Q = f(P), A = V, B = V$, in this definition. We have to show that there is a continuous function $$w: [ u, v ] \rightarrow f(P)$$ such that

Every point is $f(P)$ comes from $P$, then

(preimages of $U,V$ under $f$).

We then use the path-connectedness of $P$. We set $Q = P, A = s, B = t$, in the above definition. Then there is a continuous function $$\varphi : [ c, d ] \rightarrow P$$ such that $$\varphi (c) = s, \varphi (d) = t.$$ Now let $$w = f \circ \varphi ,$$ and let $$u = c, v = d.$$ Then

- $w$ is continuous as the composition of two continuous functions (above),
- $w(u) = f ( \varphi (u) ) = f( \varphi (c) ) = f(s) = U$,
- $w(v) = f ( \varphi (v) ) = f( \varphi (d) ) = f(t) = V. $

So function $p = w$ satisfies the above definition. $\blacksquare$

**Exercise.**

- $M(n)$ is the space of $n \times n$ matrices (over ${\bf R}$). It is identified with ${\bf R}^{n^2}$.
- $SL(n)$ is the special linear group, i.e., the space of $n \times n$ matrices with determinant equal to $1$.

Then $$SL(n) \subset GL(n)={\bf R}^{n^2}.$$ Prove that $SL(n)$ is connected.