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Can any curve in $\Bbb R^2$ (which doesn't intersect itself) be a graph of a certain equation?

In other words, if given an arbitrary curve in $\Bbb R^2$ (which doesn't intersect itself), is there a equation $f(x,y)=0$ that takes the given curve as its graph?

Edit: I was a bit confused about the definition of 'function'. What I meant was 'any equation $f(x,y)$' , such as $x^2+y^2=1$; and should not be broken down into several intervals. Sorry for the mistake.

Edit: I was asking about a continuous(?) function, as GEdgar said in the comments. I couldn't think of the right word...

Minseok Kwon
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    Let $f$ be the function such that $f(x,y)$ is 0 if $(x,y)$ is on the curve, and 1 otherwise. –  Mar 02 '19 at 11:33
  • Rahul's counterexample is good. Maybe a better question would be where the curve is "continuous" in a certain sense, and we can get a continuous function $f(x,y)$. – GEdgar Mar 02 '19 at 11:41
  • In that case just use the distance from the curve as $f(x,y)$. – Christoph Mar 02 '19 at 11:45
  • @Christoph: That only works if the curve is closed. Take the set $(0,1)\times{0}$ which probably qualifies as curve (note that the first factor here is an open interval, not a pair). Then the distance of the point $(0,0)$ (pair, this time) to that curve is $0$, but it is not element of that curve. – celtschk Mar 02 '19 at 12:16

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I think what you're asking is this:

Given a set $S \subset \mathbb{R}^2$ that represents a curve, is it always possible to find a function $f:\mathbb{R}^2 \to \mathbb{R}$ such that $S = \{(x,y) \in \mathbb{R}^2 : f(x,y)=0\}$.

The problem is that we don't have a clear definition of "curve", so we don't know much about the properties of the set $S$.

However, with some suitable restrictions on $S$, we can define a function $f$ by $$ f(x,y) = \text{(minimum) distance from } (x,y) \text{ to } S $$ This will often give you the familiar equations of various curves (like lines and circles, for example).

On the other hand, no matter what set $S$ you're given, you can define a function $f$ as suggested in Rahul's comment $$ f(x,y) = \cases{0 \quad\text{ if } (x,y) \in S \\ 1 \quad\text{ if } (x,y) \notin S } $$

Then it's certainly true that $S = \{(x,y) \in \mathbb{R}^2 : f(x,y)=0\}$, as you wanted, but the argument is somewhat circular, and we still don't know what "curve" means.

bubba
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Consider the curve in $\mathbb{R}^2$.

Graph

There exists no function $f:\mathbb R\to\mathbb R$ for which the given curve is a graph of $f$. If such a function exists then it would take three value at $x=0$ which are $0,0.5,1$, clearly, the function is not well defined.

Sujit Bhattacharyya
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  • I edited my question more clearly. I was a bit confused about the definition of 'function'. What I meant was 'any equation $f(x,y)$' , such as $x^2+y^2=1$. Sorry for the mistake..... – Minseok Kwon Mar 02 '19 at 11:31
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    Again I am confused because by means of "equation" do you mean formal equations? like as you mentioned "$x^2+y^2=1$" or "$x^{\frac{5}{4}}+y^{\frac{2}{3}}=2$"? If so then it also can not happen, for this consider the Prime counting function $\pi(x)$ in discrete case and join the points, formally $\pi(x)$ is not explicitly expressible in terms of $x$ (though you can approximate). – Sujit Bhattacharyya Mar 02 '19 at 11:42
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The graph of a function $f\colon\Bbb R\to\Bbb R$ is the set of all points $(x,y)$ such that $y=f(x)$. Note that for fixed $x$ there is exactly one point on the graph with this $x$-coordinate. Hence, any curve that contains two points of the same $x$-coordinate can not be realized as the graph of a function, so the answer is no.

If this doesn't answer your question, please edit your question to specify what you mean by

is there a function (whether implicit or explicit, etc) that takes the given curve as its graph?

Christoph
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  • I edited my question more clearly. I was a bit confused about the definition of 'function'. What I meant was 'any equation $f(x,y)$' , such as $x^2+y^2=1$. Sorry for the mistake... – Minseok Kwon Mar 02 '19 at 11:31