Path connectedness imply existence of polygonal path? Suppose $A \subset \mathbb{R}^d$ is a path connected space. $A$ is an open set. Then, for any $a,b \in A$, we can find a map $\lambda: [0,1] \to A$ and a partitiion $ \{0 = t_0 < t_1 <...< t_{m-1}<t_m = 1 \} $ of $[0,1]$ such that for all $i$, if $t \in [t_{i-1},t_i]$, then
$$ \lambda(t) = \frac{t_i-t}{t_i-t_{i-1}} \lambda(t_{i-1}) + \frac{t-t_{i-1}}{t_i - t_{i-1}} \lambda(t_i).  $$
I have no idea how to construct such a $\lambda$. 
Can someone help me? thanks
 A: Although bof's answer is correct, perhaps this is what the OP was looking for:
Let $\phi:[0,1]\rightarrow A$ be a path.For every $x \in A$, choose an open ball $U_x$ such that $x\in U_x\subset A$.
Since $\{\phi^{-1}(U_x)|x\in A\}$ forms a cover of $[0,1]$, by the Lebesgue covering lemma, there exists a partition $0=t_0<t_1<...<t_n=1$ of $[0,1]$ such that each $[t_i,t_{i+1}]$ is contained in some $\phi^{-1}(U_x)$. 
Hence each $\phi([t_i,t_{i+1}])$ is contained in some open ball. Now define $\lambda$ as $\lambda(t_i)=\phi(t_i)$ and 
$$ \lambda(t) = \frac{t_i-t}{t_i-t_{i-1}} \lambda(t_{i-1}) + \frac{t-t_{i-1}}{t_i - t_{i-1}} \lambda(t_i), \, t\in [t_i, t_{i+1}] $$
$\lambda$ is a polygonal path entirely contained in $A$. 
A: In what follows we shall show that whenever $a,b\in A$, then there exists a polygonal line inside $A$ connecting $a$ and $b$. In particular, we shall show that the set:
$$
S=\{w\in A: \text{$a$ and $w$ are connected with a polygonal line inside $A$}\},
$$
is both open and closed with respect to $A$, and hence $S$ coincides with $A$, since $a\in A$ and thus $A\ne\varnothing$ - Here we used one of the characterizing properties of connectedness: 
If $X$ is a connected space, $Y\subset X$, and $Y$ is both open and closed, then $Y=\varnothing$ or $Y=X$.
1. $\boldsymbol S$ is open. Assume that $w\in S$. Then $w\in A$, and as $A$ is open there exists an open  ball $B_r(w)\subset A$. We shall show that $B_r(w)\subset S$. If $z\in B_r(w)$, then the whole segment $[w,z]$ lies in $A$, i.e.,
$$
[w,z]=\{tw+(1-t)z: t\in[0,1]\} \subset B_r(w)\subset A.
$$ 
So if $\ell$ is a polygonal line connecting $a$ and $w$ inside $A$, then attaching $[w,z]$ in the endpoint of $\ell$ we obtain a polygonal line $\ell'$ which connects $a$ and $z$ inside $A$, and this is possible for every $z\in B_r(w)$. Thus $S$ is open.
2. $\boldsymbol S$ is closed. We need to show that if $\{w_n\}_{n\in\mathbb N}\subset S$, and $w_n\to w\in A$, then $w\in S$. Since $A$ is open, then, as before, there exists a ball $B_r(w)\subset A$, and since $w_n\to w$, there is an $n$, such that  $w_n\in B_r(w)$. Repeating the previous argument, we first observe that the segment $[w_n,w]$ is in $A$, i.e.,
$$
[w_n,w]\subset B_r(w)\subset A,
$$
and as $w_n$ is connected with $a$ with a polygonal line $\ell$ inside $A$, then so is $w$, and the polygonal line which connects $a$ and $w$ is simply the one produced when we attach at the endpoint of $\ell$ the segement $[w_n,w]$. 
A: This seems to be false. For instance, the graph of $y = \sin x$ in $\mathbb R^2$ is path-connected, but it contains no nontrivial line-segments at all, so it's hard to imagine how one might connect, say, $(0, 0)$ to $(\pi, 0)$ via a finite number of line segments, all lying in this graph. 
