I confused myself when thinking about the circle:

It can be parameterised as $$ C(t) = (\cos t , \sin t)$$

for $t \in [0,2\pi)$. This makes it clear that the circle is one dimensional. But then the circle is also defined by $x,y$ such that

$$ x^2 + y^2 = 1$$

If we try to solve this equation for $y$ then

$$ y = \pm \sqrt{1-x^2}$$

which is not a function! But if the circle was indeed one dimensional then we should be able to write it as

$$ (x,y(x))$$ which seems to be impossible. Therefore the circle is not one dimensional.

Please could someone help me resolve my confusion?

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    $\begingroup$ It's one dimensional. It can be locally written as a graph of $x$. $\endgroup$ – user99914 May 15 '15 at 5:32
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    $\begingroup$ According to your definition, a vertical line is not one-dimensional, because it cannot be written in the form $(x,y (x)) $. $\endgroup$ – Martin Argerami May 15 '15 at 5:41
  • $\begingroup$ @MartinArgerami and John: So not every one dimensional function is a function? I think I understand, thank you for helping me. It would be interesting for me to know for what conditions something one dimensional is a function ^^. Is it okay to edit the question? $\endgroup$ – Anna May 15 '15 at 5:51
  • $\begingroup$ @Anna The answer to your question is provided by the Vertical Line Test, though, as Martin's example illustrates, this depends on how the curve sits w.r.t. the axis of the candidate dependent variable. $\endgroup$ – Travis Willse May 15 '15 at 6:00
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    $\begingroup$ It doesn't quite make sense to say "every one-dimensional function is a function". (A function's domain can be $1$-dimensional, but we generally don't apply that term to the function itself.) A more precise way to put it is, "Not every curve in the $xy$-plane is the graph of some function $f$." $\endgroup$ – Travis Willse May 15 '15 at 6:02

The circle is a manifold and therefore we have to check the dimension's definition of a manifold:

A manifold $M$ has dimension $n$, when you can find for each point $p \in M$ a neighborhood $U$ of $p$ such that there is an open $V \subseteq \mathbb R^n$ and a homeomorphism $f: U \rightarrow V$

An homeomorphism is a bijective and continuous function whose inverse is also continuous. Now take the circle. You can divide it in four segments:

enter image description here

File:Circle with overlapping manifold charts.svg by User:KSmrq and User:Pbroks13 licensed under CC-BY-SA 3.0

You see, that each segment can be continuously maped to an interval (i.e. a set of $\mathbb R^1$). Thus the dimension of the manifold is 1. See also this section.

Note: You only need one information to identify a point of the circle uniquely. The angle $\phi$ in polar coordination. From this you also see, that the dimension of the circle must be one.

  • $\begingroup$ +1. I would +2 if I could, for such a complete credit line on the image. $\endgroup$ – Blue May 15 '15 at 8:30

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