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I'm trying to figure out whether or not the following set is open or closed. $$D=\{(x,y,z)\in\mathbb R^3\mid x\gt0,y\gt0,z=0\}$$

I've tried imagining it and to me, it seems like an open set, but maybe it is both open and closed. How would I determine that?

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    $\begingroup$ What do you know about open sets - what properties must they have? Likewise closed sets? "It seems like ..." doesn't refer to properties or definitions - intuition can be a guide to how you might go about the task. What do you know about closed sets and limit points, for example? Take a point in $D$ - what can you say about points in a neighbourhood of the point? (or you may have other definitions or concepts to hand) $\endgroup$ – Mark Bennet Jun 14 at 9:42
  • $\begingroup$ I was thinking about the set of all the points in the first quadrant of the X,Y plane that's why I thought it was open, but I forgot that it's a 3 dimensional set... $\endgroup$ – Counter Boosting Jun 14 at 9:48
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It is not open because it contains $(1,1,0)$ and every neighborhood if this point contains points with $z \neq 0$. It is not closed because $(\frac 1 n, \frac 1 n,0)$ is a sequence in this set which converges to a point outside the set.

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    $\begingroup$ Maybe a bit nitpicking. One has to ask: according to what topology. Of course, if your assume the natural topology of R3 then your answer is perferct. If one takes the induced topology on the subset D (and then D as a subset of that topological space) then D is open and closed. $\endgroup$ – lalala Jun 14 at 19:40
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No, it is not an open set. For instance, $(1,1,0)\in D$, but no open ball centered at $(1,1,0)$ is contained in $D$.

On the other hand, $\left(\left(\frac1n,\frac1n,0\right)\right)_{n\in\mathbb N}$ is a sequence of elements of $D$ which converges to $(0,0,0)$. But $(0,0,0)$ does not belong to $D$. What can you deduce from this?

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  • $\begingroup$ Amazing! Our answers are identical. $\endgroup$ – Kavi Rama Murthy Jun 14 at 9:40
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    $\begingroup$ Great minds think alike! $\ddot\smile$ $\endgroup$ – José Carlos Santos Jun 14 at 9:41
  • $\begingroup$ So the set is neither open or closed, thank you, both answers are great so I don't know who to award it to! :D $\endgroup$ – Counter Boosting Jun 14 at 9:46
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    $\begingroup$ @CounterBoosting I suggest that you take into account the fact that the other answer appeared before mine. $\endgroup$ – José Carlos Santos Jun 14 at 9:49
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Theorem: Let x$\in \mathbb{R^n}$ where the topology induced by the standard metric is assumed. $x\rightarrow a $ $\iff$ $x^i \rightarrow a^i$ for each $1\leq i\leq n$

Take the sequence $(\frac{1}{n},\frac{1}{n}, 0)$

Theorem 2: Let X be a metric space. $A \subset X$ is closed iff the limit of every convergent sequence in A is in A.

Observe that the components of the above sequence all converge to 0 and therefore the sequence converges to $(0,0,0)$ which is not in the set. To show that the set is open, you must find a point in the set such that no matter what radius your ball has, it is never contained in the set. The previous answers wrote $(1,1,0)$ you should prove that what they wrote indeed shows that the set is not open.

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