A curve in the complex plane is defined as a continuous function from a closed interval of the real line to the complex plane:$ z(t) : [a, b] → \mathbb{C}$, $a\leq t\leq b$.
Doubt: Why we choose closed interval to define curve in the complex plane? Can't we take an open interval to define curve?
Thanks for help
On
You buy a car to move from point $A$ to point $B$, and you define a curve for the same reason, to move from point $A=z(a)$ to point $B=z(b)$.
One could of course define a curve using an open interval, and then someone may ask why didn't we use a closed interval instead. (There is one answer to that at the very end.) One could also define a curve on a half-open interval, and (now that I wrote it) that actually turns out to be useful to provide an alternative characterization (or definition) of simply connected regions in the plane, due to Donald J. Newman. It is included and used in the textbook
Complex Analysis by Joseph Bak and Donald J. Newman (in section 8).
Springer, New York, 2010. xii+328 pp. ISBN: 978-1-4419-7287-3 (3rd edition).
They adopt a non-traditional definition of simply connected regions. Such definition allows for the simplification of some proofs, e.g. of the following results:
8.4 Theorem. Suppose $f$ is analytic in a simply connected region $D$ and $\Gamma$
is a simple closed polygonal path contained in $D$. Then $\int_\Gamma f = 0.$
8.5 Theorem.
If $f$ is analytic in a simply connected region $D$, there exists a “primitive” $F$, analytic
in $D$ and such that $F'= f .$
8.6 General Closed Curve Theorem.
Suppose that $f$ is analytic in a simply connected region $D$ and that $C$ is a smooth
closed curve contained in $D$. Then $\int_C f = 0.$
The statements of the above results are standard, but in the approach taken in the above textbook, the proofs rely on the following definition. (As usual, a region in the plane is a connected open set, and we define a simply connected region below, where $\tilde D$ denotes the complement of $D$ in the plane. Also, $d(a,B)$ denotes the distance from a point $a$ to a set $B$.)
8.1 Definition.
A region $D$ is simply connected if its complement is “connected within $\varepsilon$ to $\infty$.”
That is, if for any $z_0\in \tilde D$ and $\varepsilon > 0,$
there is a continuous curve $\gamma(t), 0\le t < \infty$
such that:
(a) $d(\gamma(t),\tilde D)<\varepsilon$ for all $t\ge 0,$
(b) $\gamma(0) = z_0,$
(c) $\lim_{t\to\infty} \gamma(t) = \infty$.
A curve $\gamma$, satisfying (b) and (c), is said to “connect $z_0$ to $\infty$.”
Note that the curve $\gamma$ is defined on the half-open interval $[0,\infty)$. The late Professor Newman adopted the above definition to provide for more elementary proofs (and perhaps also his intuition was that this definition of simply connected regions is equivalent to the usual ones).
For the proof of the equivalence of Newman's definition of a simply connected region with the more widely accepted usual definitions, one direction is easy. For the other direction and a complete proof one could see:
Joseph Bak and Strashimir G. Popvassilev,
The evolution of Cauchy's closed curve theorem and Newman's simple proof.
Amer. Math. Monthly 124 (2017), no. 3, 217–231.
https://doi.org/10.4169/amer.math.monthly.124.3.217
https://www.jstor.org/stable/10.4169/amer.math.monthly.124.3.217
Finally (just a comment that gives a good possible answer to the original question), a curve $z : [a, b] \to \mathbb{C}$ could be restricted to the smaller domain $z : (a, b) \to \mathbb{C}$. Conversely, if $z(t)$ were only defined on $(a,b)$ and $a<r<s<b$ then of course $z : [r,s] \to \mathbb{C}$ is defined. So one cannot really distinguish between the two cases easily, they are closely related, but it seems that if $z : [a, b] \to \mathbb{C}$ then we get something more, namely endpoints $z(a)$ and $z(b)$. This also allows us to compose paths $z : [a, b] \to \mathbb{C}$ and $w : [p,q] \to \mathbb{C}$ provided $z(b)=w(p)$, and that brings a lot of extra structure (the fundamental group) to study and use ( I guess I could have started with that...)
Edit: A continuous function $(a,b)\to X$ is equivalent to a continuous function $\mathbb R\to X$, because $\mathbb R$ is homeomorphic (and diffeomorphic and probably more) to an open interval. With a curve $[a,b]\to X$, we actually get something different.
Basically because we want a curve to have endpoints. Of course you could consider at open-ended curve $(a,b)\to X$, but that's not as useful.
Probably the most important reason is that it forces the (image of the) curve to be a closed and bounded set. Otherwise $z(t)=\frac 1t$ for $t\in(0,1)$ would be a curve that goes off to infinity. Even worse, you could have very complicated curves such as infinite spirals and whatnot.
As a bonus, it makes it easy to talk about the endpoints by just writing $z(a)$ and $z(b)$.