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Please Help Me:

Let $f:[a,b]\to\mathbb R$ be a continuous map. Then the image points $f(t);t\in[a,b]$ are continuously ordered according to the increasing value of $t.$

I don't understand what is meant by "continuously ordered according to the increasing value of $t$" and how can it be shown?

Of course it doesn't mean $\forall ~x,y,z\in[a,b]$ with $x< y< z,$ $|f(x)-f(y)|$$\leq |f(x)-f(z)|...(1)$ for if we consider $f:[0,2\pi]:t\mapsto \cos 2t+i\sin 2t$ then $|f(0)-f(2\pi)|<|f(0)-f(\pi)|.$ Then does it suggest $\exists$ an interval in which $1$ holds?

Ref: Churchill, Brown, Verhey [3rd Ed] enter image description here

Also how can it be guranteed that $Im~f$ always looks like a line (not necessarily straight)?

share|cite|improve this question
What are you quoting from? – Adam Saltz May 1 '13 at 3:34
@AdamSaltz: See edit – Sriti Mallick May 1 '13 at 3:41
It means $(x(t_0), y(t_0)) \lt (x(t_1), y(t_1))$ if $t_0\lt t_1$. – John Douma May 1 '13 at 4:18
@user69810: How do your define order in $\mathbb C?$ – Sugata Adhya May 1 '13 at 4:40
The ordering is only on the image-it's induced by the ordering on $\Bbb{R}$. – Kevin Carlson May 1 '13 at 5:33

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