Path-connectedness of continuous functions I want to prove that the metric space $C[0,1]$ with the metric $d(f,g) = sup_{x \in [0,1]} |f(x) - g(x)|$ is path-connected. I think I've done most of the proof, but I am not too sure about the outcome.
I simply tried the straight line path $p(t) = f(x) + t(g(x) - (f(x))$ so that $p(0) = f(x)$ and $p(1) = g(x)$. To show path-connectedness, $p \in C[0,1]$ must be satisfied, hence I need to prove that $p$ is continuous.
Let $t_0 \in [0,1]$ and $\epsilon > 0$ and assume $|t-t_0| < \delta$. Then $sup_{t \in [0,1]} |p(t) - p(t_0)| = sup_{t \in [0,1]}|t-t_0||(g(x)-f(x)| < \delta|g(x) - f(x)|$.
So if I let $\delta = \frac{\epsilon}{|g(x) - f(x)|}$ is the path p then continuous? Did I forget anything here or is this proof valid?
 A: Looks good to me, albeit a bit round-about; see my comment above for more details. As mentioned in the comments above, it's worth saying "let us fix $x$", and then show that it's true for all $x$ ($\in [0,1]$). If writing $p(t,x)$ doesn't seem nice to you, I would personally write it as $p_x(t)$ - *note that these are meaning the same thing just maybe makes it clearer what $x$ and $t$ are 'doing'.
There is something else that it's good to be aware of though. The product of two continuous functions (on $[0,1]$, say) is continuous, as is the sum. It's worth trying to show this for arbitrary functions. When you know this, you can say immediately that $p$ is continuous. Again as mentioned in the comments this makes $C[0,1]$ a vector space.

Hint: If you're struggling to show that $fg$ is continuous when $f$ and $g$ are, then consider this trick. You (hopefully) have shown that $f + g$ is. Show that $f^2$ is (should be easier as a special case). Then use the fact that $fg = {1 \over 2}((f+g)^2 - f^2 - g^2)$.

Anyway, hope this helps! =)
