2
$\begingroup$

Path connectedness seems to be defined in a topological space, but can the existence of a path be proven without using the functions of vector addition, scalar multiplication and norm ?

For example, $R^n$ is a metric space (with the Euclidean metric) and has a corresponding induced topology. But to prove that a ball (open or closed) is path connected I would construct a mapping from $ t \in [0, 1]$ to $R^n$ defining a path between $a, b \in R^n$ by $y = a + t (b - a)$ and then use the norm to show that this is continuous and within the ball. In this case (and in general) can one prove path connectedness using only metric or topological concepts ?


After comments and a couple of answers, perhaps I understand my question better. It was prompted particularly in consideration of $R^n$. It seems to be confirmed (by comments) that one cannot prove the existence of a continuous path in $R^n$ without defining a function which relies on vector addition, scalar multiplication and norm.

Path connectedness is defined for a topological space and $R^n$ induces a topology and has additional metric properties, but requires yet more properties (normed vector space) in order to prove path connectedness. This strikes me as odd because normally when a concept is defined in a mathematical structure one would expect to be able to prove various results within that structure. In this case I would have expected some, perhaps non-constructive proof, of existence of a path: apparently not.

$\endgroup$
  • $\begingroup$ You are right that don't need a norm to show topological properties of the objects used, a metric or topology would suffice. Moreover, all norms on $\mathbb R^n$ are equivalent, thus induce the same topology. Considering your mapping: How to define $b-a$ without vector addition? How to define $t(b-a)$ without scalar multiplication? $\endgroup$ – Roland Jan 16 '16 at 14:48
  • 3
    $\begingroup$ We can prove some spaces are path-connected that are not embeddable in any topological vector space, e.g. $X = \{a,b\}$ with the topology $\tau = \{\varnothing, \{a\},X\}$. So it is definitely possible to prove without vector addition, scalar multiplication and norm. $\endgroup$ – Daniel Fischer Jan 16 '16 at 14:51
  • 3
    $\begingroup$ For the example of the ball in $\mathbb{R}^n$, since that is defined in terms of the norm, I suspect it is necessary to use the norm (at least implicitly) to show that a proposed path stays within the ball. I also expect that you will have a very hard time to explicitly give an example of a non-constant continuous function $f\colon [0,1] \to \mathbb{R}^n$ that doesn't in some way use addition and multiplication. $\endgroup$ – Daniel Fischer Jan 16 '16 at 14:57
  • $\begingroup$ @DanielFischer. Thanks for feedback. The ball in $R^n$ can also be defined in terms of the metric. I was wondering if one could use metric properties of $R^n$ (e.g. a bounded sequence has a convergent subsequence) to prove path connectedness - it seems from your answer probably not. $\endgroup$ – Tom Collinge Jan 16 '16 at 15:05
  • $\begingroup$ Proving path connectivity requires the construction of a lot of paths. So you seem to be asking for new path construction techniques, ones which avoid the use of continuity of the basic arithmetic operations. I cannot even think of a way to prove path connectivity of $\mathbb{R}$ without using continuity of arithmetic. Continuity of arithmetic is so simple and fundamental, it seems very hard to avoid. $\endgroup$ – Lee Mosher Jan 16 '16 at 18:57
1
$\begingroup$

I imagine what you are asking is this:

Let $V$ be a real vector space with a topology that is compatible with addition and scalar multiplication. This means that addition $a: V \times V \to V, (x,y) \mapsto a(x,y)=x+y$ and scalar multiplication $m: \mathbb R \times V \to V, (\lambda,x) \mapsto m(\lambda,x)=\lambda \cdot x$ are continuous functions. Does this imply that $V$ is path connected?

The answer is yes.

Before the thing itself is done some simple statements:

First, let $f: X_1 \to Y_1$ and $g: X_2 \to Y_2$ be continuos functions. Define $\langle f, g \rangle :X_1 \times X_2 \to Y_1 \times Y_2$, $(a,b) \mapsto (f(a),g(b))$. Then $\langle f ,g \rangle$ is also continuous.

Second, any map from the singleton space to another topological space is continuous.

Thirdly if $\{p\}$ is the singleton space, $X$ a topological space, $X \times \{p\}$ is homeomorphic to $X$.

To see the first statement: if $U$ is open on $Y_1 \times Y_2$ then $U$ is of the form $U=\bigcup_{i \in I}U_{i,1} \times U_{i,2}$ with $U_{i,1}, U_{i,2}$ open on $Y_1, Y_2$ because sets of that form are a basis of the product topology. Note that $$\langle f, g\rangle^{-1}(\bigcup_{i \in I}U_{i,1} \times U_{i,2})=\bigcup_{i \in I}\langle f, g\rangle^{-1}(U_{i,1} \times U_{i,2})=\bigcup_{i \in I}f^{-1}(U_{i,1})\times g^{-1}(U_{i,2})$$

Since $f,g$ are continuous $f^{-1}(U_{i,1})\times g^{-1}(U_{i,2})$ is open in $X_1 \times X_2$ and $\langle f, g\rangle^{-1}(U)$ is then open for any open $U$.

The second and third statements are trivial.

Now define for any $x \in V$ $s_x: \{p\} \to V, p \mapsto x$. Then define $\gamma_x :=m(\langle id, s_x \rangle) : [0,1]\times\{p\} \to V$. Because $[0,1] \times \{p\}$ is the same as $[0,1]$ I will just throw away the dependence of construction on the singleton set from now on. As a composition of various continuous maps $\gamma_x$ is also continuous, the specific form is $\gamma_x(t)= t \cdot x$, so $\gamma_x$ is a continuous path connecting the points $0$ and $x$.

You can either use that two points being connected by paths is an equivalence relation, or define $\gamma_{x,y}:=a(\langle \gamma_x, \gamma_y \circ i \rangle$ with $i:[0,1] \to [0,1], t \mapsto 1-t$ to find $\gamma_{x,y}(t) = t\cdot x + (1-t) \cdot y$ is a continuous path connecting $x$ and $y$.

This is then a general statement that requires only a compatibility of the vector space structure with the topology. If no such compatibility is required, the property does not need to hold, the topology has now no connection with the vector space structure apart the set $V$ being limited in what cardinality it may have (basically no finite sets apart from $\{0\}$ and no countable sets are allowed).

$\endgroup$
-1
$\begingroup$

A metric space might not be connected (and therefore path connected). Consider for example the space $(0,1) \cup (2,3)$ equipped with the standard real distance.

One important property of a Normed Vector Space is that it is convex, hence path connected.

I don't know if it precisely answer what you had in mind...

$\endgroup$
  • 1
    $\begingroup$ Thanks for feedback. Not really my question though. Perhaps start from the example - can a ball in $R^n$ be proved path connected purely using metric and topological properties (i.e. without using vector addition and scalar multiplication to construct a path function) ? $\endgroup$ – Tom Collinge Jan 16 '16 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.