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I suppose to more formally characterize the question more formally, why are all points of the set $\{ (x,y) \mid F(x,y) = 0 \}$ always boundary points (and I believe also never isolated points) in the standard topology of $\mathbb{R}^2$ -- for simplicity's sake, say in the case where $F(x,y) \in \mathbb{R}[x,y]$ (i.e. polynomials in $x,y$ with coefficients in $\mathbb{R}$)?

I imagine there are standard proofs of such a proposition (and certainly more general than this) in algebraic geometry, but I have not studied algebraic geometry, and was wondering if there were any elementary proofs of this, only relying on basic point-set topology, algebra, real analysis, etc... I'm also looking for a relatively intuitive explanation/proof, if possible. Maybe the most intuitive reason why this is true is because of the implicit function theorem, but I wonder if there are other, still intuitive, elementary proofs of this.

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  • $\begingroup$ boundary points of what? as for isolated points, what about f(x,y) = x^2 + y^2? $\endgroup$ – zhw. May 14 '16 at 18:16
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    $\begingroup$ My educated guess is that Implicit Function Theorem is what you are looking for. Note that in special cases you do get an isolated point - $F(x,y)=(x-1)^2+(y-2)^2$ vanishes only at $P=(1,2)$. Sometimes you get the empty set - $F(x,y)=1+x^2+y^2$. $\endgroup$ – Jyrki Lahtonen May 14 '16 at 18:18
  • $\begingroup$ Sometimes the zero loci are not smooth curves, even for smooth functions $F$; consider, e.g., $F(x, y) := xy$, which is singular at the origin. $\endgroup$ – Travis Willse May 14 '16 at 18:21
  • $\begingroup$ @zhw Boundary points of the set itself. The intuition is, if our set had a "filled in area", there would be non-boundary points, but in the case of a one dimensional curve, every neighborhood of every point contains points not in the set, and points is the set, and hence is a boundary point. $\endgroup$ – Nathan BeDell May 14 '16 at 18:27
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Suppose $p(x,y)$ is a polynomial that is not identically $0.$ Let $Z$ be the zero set of $p.$ Then $Z$ is closed, hence $Z = \text { int } Z \cup \partial Z.$ Suppose $\text { int } Z $ is nonempty. Then there is an open disc $D(a,r) \subset Z.$ Then for any nonzero $v\in \mathbb R^2,$ the function $p_v(t) = p(a+tv)$ is a one variable polynomial in $t$ that vanishes in a neighborhood of $0.$ Thus $p_v$ vanishes identically. Since this is true for all $v,$ $p \equiv 0,$ contradiction.

This result also holds for any real analytic function $f(x,y)$ on $\mathbb R^2.$ The proof is basically the same. Beyond this, there are plenty of functions in $C^\infty(\mathbb R^2)$ for which the zero set has non-empty interior.

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If "intuitive" is what you want, then think of what the graph of a function of two variables looks like. Hills and valleys. If you have $F(x,y) = \text{specified constant}$, then the value of the constant tells you to what depth you fill fill the world with water, and then the curve is the coastline. Why is the coastline a curve? That is the question.

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