Is a Normed Vector Space Necessary to Prove Path Connectedness? 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. 
 A: 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).
A: 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...
