Is there a connected subset of $\mathbb{R}^n$ that is not complete? Thanks in advanced
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Is $\Bbb{Q}$ complete? – Race Bannon Aug 07 '15 at 22:54
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2@Race: Is $\Bbb Q$ connected? – Asaf Karagila Aug 07 '15 at 22:55
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@AsafKaragila Normal Human said any subset, not any connected subset. – Race Bannon Aug 07 '15 at 22:56
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5Any open and bounded subset under the usual topology isn't closed: therefore there are points on the boundary which are the limits of sequences in the set, but the boundary points are not contained in the set. – Mnifldz Aug 07 '15 at 22:57
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$\mathbb{Q}$ is not complete nor connected. Any closed subset of $\mathbb{R^n}$ is complete – KioskeMamo Aug 07 '15 at 22:58
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possible duplicate of Connectedness of sets in the plane with rational coordinates and at least one irrational – Tomasz Kania Aug 07 '15 at 23:12
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1@Tomek: It most definitely is not a duplicate of that question. – Brian M. Scott Aug 08 '15 at 00:01
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1I don't understand what is the fuss on this question, relating to its downvotes/upvotes. This doesn't seem homework-seeking at all: it seems like a confused person who just got introduced to Analysis, imagining things that are "connected" should have no "separation", hence be "complete". This is exactly why I tried to answer in a detailed but simple form. Is the question trivial? Yes, one could argue so. But why isn't this same behaviour being reproduced in this question, for instance?: http://math.stackexchange.com/questions/1388242/function-grows-slower-than-lnx – Aloizio Macedo Aug 08 '15 at 00:35
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I have posted about the simplest possible answer. – ncmathsadist Aug 08 '15 at 01:46
2 Answers
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Take $A:=(0,2) \times \{0\}\times \{0\}... \times \{0\}$, and the sequence $x_n=(\frac{1}{n},0,0,...,0)$
It is clearly cauchy, but it doesn't converge. Why? Suppose otherwise, that $x_n \rightarrow x \in A$. Since we are taking the induced metric, this would imply that $x_n \rightarrow x$ also in $\mathbb{R}^n$. But note that $x_n \rightarrow 0$ in $\mathbb{R}^n$. Since limits are unique in a metric space, we arrive at a contradiction.
Now, $A$ is connected, since it is the image of $(0,2)$ under the continuous map $i: (0,2) \hookrightarrow \mathbb{R}^n$
Aloizio Macedo
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I won't keep this answer just to get a positive score, if it is inappropriate... therefore, I must ask: why the downvotes? – Aloizio Macedo Aug 07 '15 at 23:31
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1There's a lot of weird downvoting today. Jupiter must be aligned with Mars. – zhw. Aug 07 '15 at 23:32
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