A convex subset of normed vector space is path-connected Let $(N, \|\;\|)$ be a normed vector space and $(X,\tau)$ a convex subset of $(N,\|\;\|)$ with its induced topology. Show that $(X,\tau)$ is path-connected, and hence also connected.
What I have done so far is Let $a, b \in N$, then construct $X = \{x: x = at+ (1-t)b, 0\le t \le 1\}$. Let $f$ be a function from $[0,1]$ to $X$ with $f(0) = a$ and $f(1) = b$.
I guess if I define $f(t) = at+(1-t)b$, but can I do this without knowing the form of $\| \; \|$? 
 A: Let $x\in X$ and $r>0$, and consider the open ball $B(x;r)=\{y\in X:\|x-y\|<r\}$. For any $y,z\in B(x;r)$ and $t\in(0,1)$, we have
\begin{align}
\|ty + (1-t)z - x\| &= \|t(y-x) + (1-t)(z-x)\|\\
&\leqslant t\|y-x\| (1-t)\|z-x\|\\
&<r,
\end{align}
so $B(x;r)$ is convex. Write $f(t) = t(a-b) + t$, then it is clear that $f$ is an affine function. So if $B(x;r)$ is an open ball in $X$ and $u,v\in f^{-1}(B(x;r))$, then $f(tu + (1-t)v)\in B(x;r)$ for any $t\in(0,1)$, from which it follows that $f^{-1}(B(x;r))$ is convex. In particular, for any distinct $x,y\in X$, the preimage of the line segment connecting $x$ and $y$ (excluding the endpoints) $f^{-1}(\{ tx+(1-t)y:0<t<1\}=(0,1)$ is open in $[0,1]$. If $y\in\delta B(x;r)$ then there is a unique point $z\in\delta B(x;r)$ with $\|y-z\|=1$. It follows that $$B(x;r) = \bigcup_{\{y:\|x-y\|=r\}}\{ty+(1-t)z:0<t<1\} $$ is the union of open line segments, and hence $$f^{-1}(B(x;r))=(0,1).$$ It follows that $f$ is continuous and therefore $X$ is path-connected.
By the way, convex sets are actually simply connected, as if $f$ and $g$ are paths in a convex set $X$ then the map $H:[0,1]^2\to\mathbb R$ defined by $$H(s,t)=(1-t)f(s) + tg(s)$$ is a path homotopy.
