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Is $M=\left \{(x,y)\in (0,\infty )\times\mathbb{R} : y=\sin(\frac{1}{x}) \right \}$ a closed set in space $((0,\infty )\times\mathbb{R} ,\rho_{e})$, $\rho_{e}$ - Euclidean metric ? I think that open set could be built "around" function $y=\sin(\frac{1}{x})$ on domain $(a,\infty)$ where $a>0$ with sum of open balls, but I don't know if it's true, and I really don't have an idea what to do in the main case of - that is for $a=0$, mayby by induction ?

This is much less important : And what happens (in short) in the case of other metrics or when we change $(0,\infty )\times\mathbb{R}$ into $[0,\infty )\times\mathbb{R}$ in definition of $M$ and metric space ($M$ have to be expanded to $x=0$ case, and it seems that $\forall\ {a\in (0,1)}\Bigl(\ \exists\ {b_{n}\subset M}\text{ such that }\lim\limits_{n\to \infty }b_{n}=(0,a)\Bigr)$, so $M$ is going to be not closed because there exist seuences with limit points outside set $M$ for every value $y$ of $M$ for $x=0$ which we choosed).

$\rho_{e}$ - Euclidean metric

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What is $\rho_{\epsilon}$? – copper.hat May 14 '12 at 0:04
There seems to be a spurious $t$ in the title? – joriki May 14 '12 at 0:06
$\rho_{e}$ - Euclidean metric, sorry – Qbik May 14 '12 at 14:30
up vote 2 down vote accepted

This is contingent on what $\rho_e$ is, but let me give you an answer that may help you. Consider $\mathbb{R}^2$ in the standard topology induced by the euclidean norm. The set

$X=\bigl\{(x,y)\in\mathbb{R}^2:x>0,y=\sin(1/x)\bigr\}\cup\bigl\{(0,y)\in\mathbb{R}^2:-1\leq y\leq 1\bigr\}$

is closed in $\mathbb{R}^2$, and so your set $M$ is closed in $(0,\infty)\times\mathbb{R}$ under the subspace topology, as the intersection of $X$ and $(0,\infty)\times\mathbb{R}$. The subspace topology, though, is also induced by the euclidean norm, restricted to $(0,\infty)\times\mathbb{R}$.

Now, depending on what $\rho_e$ is, you may be able to use a similar argument.


You are correct in your conjecture that the given set $M$ is no longer closed as a subspace of $\mathbb{R}\times[0,\infty)$, and that is certainly a fair way to justify it--note that we cannot actually change $(0,\infty)$ to $[0,\infty)$ in the definition of $M$, though (since $1/0$ is not defined), unless we further explicitly state that we require $x>0$.

As for the "other metrics" bit, that really does depend on which metric you choose, as I stated above. Obviously, the answer will be the same for any metric that induces the same topology as $\rho_e$, but with an arbitrary metric, that may or may not be the case. For example, consider the metric

$d(x,y)=\begin{cases}0 & x=y\\1 &x\neq y.\end{cases}$

This is called the discrete metric, and introduces the discrete topology on the plane. In the discrete topology, every subspace of $\mathbb{R}^2$ is both open and closed, so $M$ is open and closed as a subspace of any subspace of $(\mathbb{R}^2,d)$ that contains it. There may be yet other metrics in which $M$ fails to be closed even in the initial context, though none spring to mind.

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I shouldn’t be at all surprised if $\rho_e$ is the Euclidean metric. – Brian M. Scott May 14 '12 at 3:16
Don't ruin the suspense... – copper.hat May 14 '12 at 6:48
$\rho_{e}$ - Euclidean metric, sorry – Qbik May 14 '12 at 14:31
Quite alright, @Qbik. In that case, it should answer your question, unless of course there's some detail(s) you'd like me to elucidate further? – Cameron Buie May 14 '12 at 15:10
Ah! I see your edit now. I will edit my answer accordingly. – Cameron Buie May 14 '12 at 15:14

Another way of directly seeing that $M$ is closed in $(0,\infty)\times \mathbb{R}$ is by considering $f:(0,\infty)\times \mathbb{R}\to \mathbb{R}$ so that $f(x,y)=\sin(\frac{1}{x})-y$ (which is clearly continuous) and observing that $M=f^{-1}\{0\}$ whence $M$ is closed as a preimage of a closed set under a continuous function.

The notion of $M$ cannot be extended to $[0,\infty)\times \mathbb{R}$ as it stands since $\sin(\frac{1}{0})$ is not defined, but you can observe that its closure in $\mathbb{R}^{2}$ contains all points $(0,y)\in\mathbb{R}^{2}$ where $y\in [-1,1]$, since every open ball around such point contains a point $(x,\sin(\frac{1}{x}))$ for some $x>0$.

And to the question in your first paragraph: note that Induction is a proof over the set of natural numbers, not reals.

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I like your answer also, but Cameron Buie was first. – Qbik May 14 '12 at 20:54
Sure; which ever you prefer :-) – T. Eskin May 15 '12 at 6:35

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