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This is a question following this one. I pictured all the points of:

$$C = \{x\times 0 \mid 0<x<1\}$$

as a subset of the ordered square. (the question asks me to find the closure)

I calculated its limit points, which are $[0,1)\times 1$, andthe points of $C$, which are $(0,1)\times 0$. In order to calculate the closure (which is what the exercise asks, I need to unite the limit points with the set). Brian told me that $\langle 1,0\rangle$ is also a limit point, but I can't see why:

Look at the image below:

If I take an open set around $\langle 1,0\rangle$, there's no intersection with $C$, which are the points that lie on the ground, that is, the points on $(0,1)\times 0$

enter image description here

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Guerlando OCs
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  • Standard notation is $\langle x,y\rangle$, not $<x,y>$. I edited accordingly. $\qquad$ – Michael Hardy Sep 13 '16 at 22:25
  • I agree with you: I don't think $\langle 1, 0 \rangle$ is in the closure of $C$, so maybe there has been a misunderstanding in the discussion of the other question. However this isn't a new question and I think you should be following it up by comments and edits of the earlier question. Hence (with apologies) I vote to close this question. – Rob Arthan Sep 13 '16 at 22:43
  • @RobArthan I just didn't want to end up in an infinite discussion, as it's recommended to not be done, by the forum – Guerlando OCs Sep 13 '16 at 22:44
  • Guerlando:I understand: the policy is far from clear. I think an extended discussion in one place is better than raising a new question just to continue discussion about one aspect of one question. Ask on Meta-MSE if you want more clarification on the policy. – Rob Arthan Sep 13 '16 at 22:51

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