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I'm absolutely not sure if I got this right (I don't think I did), so it would be great if someone could check if my "proof" makes sense or not. I have to determine whether the set $$B := \mathbb{R}^2 \setminus \left\{\left(x, \sin\left( \frac{1}{x}\right)\right) : x \in \mathbb{R} \setminus \{0\} \right\}$$ is open or not.

What I did so far:

If $B$ is open, then the set $B^{\complement} := \left\{\left(x, \sin\left( \frac{1}{x}\right)\right) : x \in \mathbb{R} \setminus \{0\} \right\}$ should be closed.

Let's consider a singleton $b_n:=\{(x_n, \sin(\frac{1}{x_n}))\}$ where $x_n \ne 0$ and $n \in \mathbb{N}$.

What I would like to do is to show that each such singleton is closed, so that each set $\mathbb{R}^2 \setminus \{(x_n, \sin(\frac{1}{x_n}))\}$ is open, and therefore the union of all of them is open, that is: $B$ would be open. I'm not sure if it makes sense or not.

To show that a singleton is closed, we could let $y_n := (x_n, \sin(\frac{1}{x_n}))$ and let $\{y_n\} \subseteq b:=(x, \sin(\frac{1}{x}))$ be a convergent sequence (not sure if that's possible though, since $\sin(\frac{1}{x})$ is not well-known to be convergent). If that could be a thing, we would have $y_n = (x, \sin(\frac{1}{x}))$ (because $y_n \subseteq b$, with $b$ containing the single point $(x, \sin(\frac{1}{x}))$) and so $y_n \rightarrow (x, \sin(\frac{1}{x})) \in b$. Therefore, each $b_n$ would be closed and the union of all $\mathbb{R^2} \setminus \{(x_n, \sin(\frac{1}{x_n}))\}$ with $x_n \ne 0$, $n \in \mathbb{N}$ is open.

...Now that I read all this again, I feel like this is terribly wrong. Therefore, it might be that $B$ is not open. But if that's the case, I'm stuck at finding any counterexample of an open ball that would not be in $B$... Any help would be greatly appreciated!

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  • $\begingroup$ Hint: look at the origin. $\endgroup$ – Arthur Apr 20 '17 at 9:08
  • $\begingroup$ @Arthur Yes, $0$ is not included... but what should I conclude from that? I'm stuck (and not good at all with limits, obviously). $\endgroup$ – justdoit Apr 20 '17 at 9:15
  • $\begingroup$ "... and therefore the union of all of them is open, that is: $B$ would be open." You're right that the union of all the $\mathbb{R}^2 \setminus \lbrace (x_n, \sin(\frac{1}{x_n})\rbrace$ would be open, but this union is equal to $\mathbb{R}^2$, not $B$. $B$ is the intersection of all these sets, but an infinite intersection of open sets need not be open. $\endgroup$ – Bib-lost Apr 20 '17 at 9:15
  • $\begingroup$ The origin is in $B$. What about open balls around the origin? $\endgroup$ – Arthur Apr 20 '17 at 9:19
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    $\begingroup$ @Bib-lost Thanks a lot for the help and all the answers, I'm going to study and read all this carefully :) $\endgroup$ – justdoit Apr 20 '17 at 9:46
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Now that I read all this again, I feel like this is terribly wrong.

Your feeling is correct. :P Singletons are always closed because $\mathbb{R}^n$ is T1. That's one thing. The other is that $B$ is not a union of complements of singletons. Actually if $x, y$ are two different points, then $(X\backslash\{x\})\cup(X\backslash\{y\})=X$. So your $B$ is actually an intersection of complements. However infinite intersections need not be open.

Now your first step was correct. It is enough to show that the graph of a continous function is closed. For that see this:

Why is the graph of a continuous function to a Hausdorff space closed?

Now the last problem is: the complement of the graph should be open in what? In codomain? In $\mathbb{R}^2$?

Denote by $f(x)=\sin(\frac{1}{x})$. What the article above proves is that the graph

$$Gr(f)=\{(x,f(x))\ |\ x\in\mathbb{R}\backslash\{0\}\}$$

is closed in $Y=\mathbb{R}\backslash\{0\}\times\mathbb{R}$. So the complement $Y\backslash Gr(f)$ is open in $Y$. But $Y$ is open in $\mathbb{R}^2$ and thus the complement of graph is also open in $\mathbb{R}^2$.


However note that $B\neq Y\backslash Gr(f)$ (thanks @Bib-lost for realizing that). They are almost equal, note that the graph is defined in $\mathbb{R}\backslash\{0\}\times\mathbb{R}$. Thus

$$B=(Y\backslash Gr(f))\cup\{(0,r)\ |\ r\in\mathbb{R}\}$$

This set is not open in $\mathbb{R}^2$. That's because there is no open neighbourhood of $(0,0)$ which does not intersect the graph $Gr(f)$.

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  • $\begingroup$ $Gr(f)$ is closed in $Y$, but definitely not in $\mathbb{R}^2$, as ${0} \times [-1, 1]$ lies in its closure. $\endgroup$ – Bib-lost Apr 20 '17 at 9:18
  • $\begingroup$ @Bib-lost And where did I say that $Gr(f)$ is closed in $\mathbb{R}^2$? $\endgroup$ – freakish Apr 20 '17 at 9:19
  • $\begingroup$ Just realized how stupid it was to say that $B$ is the union of singletons... Let's say it's because I'm tired as hell :) I'm going to check that article, thanks a lot! $\endgroup$ – justdoit Apr 20 '17 at 9:19
  • $\begingroup$ @freakish You didn't, I'm afraid I do not understand your last sentence. What do you mean by "thus the graph is also open in $\mathbb{R}^2$? $\endgroup$ – Bib-lost Apr 20 '17 at 9:20
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    $\begingroup$ @Bib-lost Ah, you're right. Thanks, I've updated the answer. $\endgroup$ – freakish Apr 20 '17 at 9:30
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The sequence

$$ (\frac{1}{\pi}, 0), (\frac{1}{2\pi}, 0), (\frac{1}{3\pi}, 0), (\frac{1}{4\pi}, 0), \ldots $$

is not contained in $B$, but its limit $(0, 0)$ is. Hence, $B$ is not open in $\mathbb{R}^2$. As freakish pointed out, $B \setminus (\lbrace 0 \rbrace \times \mathbb{R})$ is however open in $\mathbb{R} \setminus \lbrace 0 \rbrace \times \mathbb{R}$.

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B^c is not closed because (0,0) is not in B^c and every open nhood of (0,0) intersects B^c. It is not open because the point p = (1, sin 1) is in B^c but no nhood of p is contained in B^c. Exercise. What is the closure of B^c? What is its interior?

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