# Is this set open or closed (or both?)

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?

• 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) – Mark Bennet Jun 14 at 9:42
• 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... – Counter Boosting Jun 14 at 9:48

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.

• 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. – lalala Jun 14 at 19:40

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?

• Amazing! Our answers are identical. – Kavi Rama Murthy Jun 14 at 9:40
• Great minds think alike! $\ddot\smile$ – José Carlos Santos Jun 14 at 9:41
• 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 – Counter Boosting Jun 14 at 9:46
• @CounterBoosting I suggest that you take into account the fact that the other answer appeared before mine. – José Carlos Santos Jun 14 at 9:49

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.