I am working on a homework question that gives $A \subseteq X$ and three points $x, y, z \in A$. Suppose also that there are continuous functions $f, g : [0,1] \to X$ with $f([0,1]) \subseteq A$, $f(0)=x$, $f(1)=y$ and $g([0,1]) \subseteq A$, $g(0)=y$, $g(1)=z$. I am asked to show that there is a function $h:[0,1] \to X$ with $h([0,1]) \subseteq A$, $h(0)=x$, and $h(1)=z$.

I think I have proved that $h(0)=x$ and $h(1)=z$, but I am not sure how to prove $h([0,1]) \subseteq A$. So far, I have attempted to have $h=f+g$ and use the triangle inequality to show that $d(x,z) < d(f(0),g(1))$ and say that this implies $h([0,1]) \subseteq A$. To be honest, I am not sure that this is the correct way to do this. Any help or advice would be appreciated.


Imagine that the parameter in each of the paths measures time, so for instance, from time $t = 0$ to time $t = 1$, a point moves from $x$ to $y$ according to the function $f$, all the while staying within $A$.

The function $g$ parametrized a path from $y$ to $z$ during that same time interval. But what if we delayed the start of $g$ so that it ran from $t = 1$ to $t = 2$? Then, we would have a point moving from $x$ to $y$ during $[0, 1]$ and from $y$ to $z$ during $[1, 2]$. Here's a piecewise formula for such a path, where I have deliberately chosen to call the variable $s$: $$ k(s) = \begin{cases} f(s) & 0 \le s \le 1 \\ g(s - 1) & 1 \le s \le 2 \end{cases} $$

This function almost satisfies all your criteria. Do you see why $k([0, 2]) \subseteq A$, for example? The one issue with $k$ is that although it gets the point from $x$ to $z$, it takes too long! There's a simple solution: run the filmstrip of $k$ at double speed.

Let $s = 2t$. The resulting function $h: [0, 1] \to A$ has formula $$ h(t) = \begin{cases} f(2t) & 0 \le t \le \frac{1}{2}\\ g(2t - 1) & \frac{1}{2} \le t \le 1 \end{cases} $$ does the job nicely.

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    $\begingroup$ Thanks for your detailed reply. This function makes sense - to formalize that $h$([0,1])$\in$$A$ then, I would simply need to state that $f$([0,1]) and $g$([0,1]) are both in $A$ correct? $\endgroup$
    – kpz
    Nov 25 '13 at 8:23
  • $\begingroup$ That's right. All the values of $h$ are values of either $f$ or $g$, which live in $A$. $\endgroup$ Nov 25 '13 at 14:52

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