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Given the graph of the curve $y=\frac{1}{x}$, can we determine if the curve is closed or open in $\mathbb{R}^{2}$ with the standard topology?

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    $\begingroup$ First, do you think it could be open? Visualize what the curve looks like and what open balls look like in $\mathbb{R}^2$. Also, remember that to show something is closed, you can show that its complement is open. What would be the complement of a curve? $\endgroup$ – Ryan Sullivant Apr 13 '14 at 19:54
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    $\begingroup$ Also, do you think this curve has any limit points in the plane that are outside of the curve itself? You can restate this as: can you find a sequence of points on this curve that converge to some point off of the the curve? If not, then the curve contains all of its limit points and so is closed. $\endgroup$ – MPW Apr 13 '14 at 20:09
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    $\begingroup$ @RyanSullivant: Your first question is quite misleading. Closed is NOT not open. $\endgroup$ – Ted Shifrin Apr 13 '14 at 20:18
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    $\begingroup$ "Sets are not doors" :) $\endgroup$ – Riccardo Apr 13 '14 at 20:19
  • $\begingroup$ @RyanSullivant, i'm not sure how to visualize or interpret the complement of a curve. Could you expand upon that. Also, I do realize sets are not doors. :P $\endgroup$ – Pubbie Apr 13 '14 at 21:00
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The curve is the inverse image of a closed set by a continuous function: $$g(x,y)=xy,$$ $$\text{graph}=g^{-1}(\{1\}).$$

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  • $\begingroup$ How can I prove this? $\endgroup$ – Pubbie Apr 14 '14 at 1:22
  • $\begingroup$ Is a standard result. Prove first that the inverse image of an open set is open and take complements. Inverse images are well-behaved with set-theoretic operations. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 14 '14 at 6:41
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Perhaps the easiest way to show closedness of $$G=\{(x,\ 1/x)\mid x\ne 0\}\subset\Bbb R^2$$ is to note that it is the preimage of $\{1\}$ under the map $$\Bbb R\times\Bbb R→\Bbb R,\qquad (x,y)\mapsto x\cdot y$$ and this map is continuous and $\{1\}$ is closed.

Alternatively, note that $G=G(f)$, the graph of the continuous map $$f:\Bbb R\setminus\{0\}\to\Bbb R,\qquad x\mapsto f(x)=1/x$$ and if $f:X→Y$ is continuous and $Y$ is Hausdorff, then $G(f)$ is closed in $X×Y$. This, however, only gives closedness of $G$ in $\Bbb R\setminus\{0\}×\Bbb R$. You still had to show that no point in $\{0\}×\Bbb R$ is in the closure of $G$.

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Note that $y = \frac{1}{x}$ has a vertical asymptote at $x = 0$, and really consists of two disjoint curves (in quadrants one and three). Picture the plane $\mathbb R^2$ with these two disjoint curves deleted. This is the complemement of $y = \frac{1}{x}$. Try to prove that this complement is open by showing that each of its points is an interior point. (I would take an $\epsilon$-neighborhood about each point $x$ in the complement, such that $\epsilon$ is less than the shortest distance from $x$ to the curve.)

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