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Let $I$ be an open interval of $\mathbb R$.

Does there exist a continous function $f : I \rightarrow \mathbb R^n$ such that $(\mathrm{range} f) \cap \mathbb Q^n = \emptyset$?

In general, what I'm really wondering is if it's possible to have a continuous curve completely 'miss' a set of values which is dense in its codomain. (Something besides the trivial case when $f$ only ever takes on a single value)

In this case, $\mathbb Q^n$ is dense in $\mathbb R^n$, and I wonder if it's possible to craft a continuous function which avoids any members of this set. If so, this gives rise to some interesting ideas.

The answer when $n = 1$ is shown to be negative by the intermediate value theorem. But I'm wondering about more general spaces.

Myridium
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  • One and two and three essential duplicates. And of course this which itself is a duplicate, with many links. – Asaf Karagila May 01 '15 at 12:58
  • @AsafKaragila I haven't learnt the concepts of connectedness or some of the notation, and so I have no way of knowing that those questions are actually asking the same thing. Mine is a separate question stated in an entirely different way. Two questions may ultimately have the same answer, but that does not mean that they were duplicate questions. If I knew in the first place that the answer would be the same, why would I ask the question? – Myridium May 01 '15 at 13:01
  • So you're saying that if I took your question, and then rewrote the definition of continuity explicitly, without using the term continuity; and also expanded the definition of rational and irrational without using those terms... it would be a completely different question? No, sorry, I don't buy it. – Asaf Karagila May 01 '15 at 13:03
  • @AsafKaragila I suppose not. But if you don't give any leeway with this kind of thing, then the only people who could ask questions would be those with such highly developed mathematical expression that they could read an existing proof or make one themselves. I suppose that's just the way Math.SE works. – Myridium May 01 '15 at 13:15
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    I don't know why you're getting all that defensive. I merely pointed out that this question, essentially, was asked before several times. Did I even vote to close your question? – Asaf Karagila May 01 '15 at 13:20
  • @AsafKaragila If you think this is a duplicate, then I think you should vote to close it. – Myridium May 01 '15 at 13:35

2 Answers2

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How about $f(x) = (\sqrt 2,x,x,\ldots,x)$?

Umberto P.
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  • Such a simple example! This does answer my question; I am just considering some generalisation that is more in the spirit of my underlying question, and may alter my post soon to accommodate it. – Myridium May 01 '15 at 13:04
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For $n=2$ we have for example $$f(x)=(x,\sqrt 2x+\sqrt3(x+1))$$ or $$f(x)=(x,\sqrt2+e^x)$$

ajotatxe
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