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What is the definition of the curve?

a. Is the image of $x^2+y^2=1(x\neq0)$ a curve?

b. Is a point a curve?

Here are the definitions I found in Wikipedia that may help.

  1. A curve is a topological space which is locally homeomorphic to a line.

  2. Let I be an interval of real numbers (i.e. a non-empty connected subset of $R$). Then a curve is a continuous mapping $\gamma:I\to{}X$, where X is a topological space.

Two objects are homeomorphic if they can be deformed into each other by a continuous, invertible mapping. Right? But the second definition doesn't mention about the invertible.

Where can I get a rigorous definition of curve? Or in which topology textbook can I find the answer?

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$x^2+y^2=1$ with $x\neq0$ would be a circle with two holes, right? For the second point: is a point homeomorphic to a line? – Timmy Turner Dec 2 '11 at 6:21
$x^2+y^2=1$ is an ellipse, ellipse is a conic section and a point is also a conic section... – VelvetThunder Dec 2 '11 at 6:25
@Timmy The image of $x^2+y^2=1$ with $x\neq0$ would be a circle with two holes. What makes me confused is the definition of the curve. – Charles Bao Dec 2 '11 at 6:36
We can consider a simpler problem. Is the image of $y=\dfrac{x^2-4}{x-2}$ a curve? – J. M. Dec 2 '11 at 6:42
@J.M. then can you get a continuous, invertible mapping? – Charles Bao Dec 2 '11 at 6:46
up vote 5 down vote accepted

I think the second definition that wiki gave is the most intuitively. Namely, one can think of a curve in a topological space $X$ as being a "path" traced out continuously, where the trace is indexed by time. From this it immediately follows that the a) is not a curve. One can't take a stroll in $\mathbb{R}^2$ and hope to jump from the above the $x$-axis to below. Similarly, a point is a curve. It corresponds to the path where we go nowhere. This is automatically continuous since, for us to screw up continuity (make a jump), we'd have to go somewhere!

More mathematically, a) cannot be a curve since any curve is the continuous image of the connected space $I=[0,1]$ and thus itself is connected. Since a) is not connected, it can't be a curve. b) is a curve since any constant map is continuous.

Perhaps intriguing though to you is the notion of $1$-manifolds which correspond to spaces that "locally look like (non-trivial) curves" (your definition 1.). In this case a) is a $1$-manifold since near any point a) looks like a bending/scaling of $I$ whereas b) is not since you can't scale $I$ down to a point! More rigorously, you can take the upper and lower semi-circles and paramaterize them by the maps $t\mapsto e^{\pi i t}$ for $t\in(0,1)$ and $t\mapsto e^{-\pi i t}$ $t\in(0,1)$.

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+1 for "taking a stroll in $\mathbb{R}^2$" :) – Bruno Joyal Dec 2 '11 at 7:10
@Alex But in definition 1, homeomorphic requires the mapping not only continuous but also invertible. The maps t↦eπit for t∈(0,1) and t↦e−πit t∈(0,1) seems not continuous nor invertible. – Charles Bao Dec 3 '11 at 2:38
Hmm? What makes you think that this is true. It's continuous because it's the exponential map! It's invertible since the fundamental period is $2$! – Alex Youcis Dec 3 '11 at 3:38
@AlexYoucis I'm not familiar with geometry terms about exponential map and fundamental period. But I know that if the mapping is invertible, it should be bijective. Right? Can you recommend some geometry textbook? Thanks. – Charles Bao Dec 3 '11 at 10:25

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