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Is there a one-one onto continuous function $f:[0,1]\rightarrow[0,1]^2$?

I was trying to prove that there is no such function, but failed.

Any suggestions?

  • 3
    It might be a good idea to post what you've tried :)2012-01-21
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    one - one just means injective. There are many such maps. Only if you in addition require the map to be onto, then there is no such map.2012-01-21
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    I just noted that my comment is not quite correct. There my be, of course, many other condition you could impose on $f$ which make it impossible find such a function. But the one you have in mind is probably surjectivity :-)2012-01-21
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    Any 1-1 onto continous function from $[0,1]$ to $[0,1]^2$ would be a homeomorphism since we are dealing with compact spaces. (That is the inverse would automatically be continuous.) Can you think of a good reason the two spaces are not homeomorphic? What happens when you remove a point from each space?2012-01-21
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    I didn't learn topology, it is a question from calculus 2 course (under the subject of functions from R^n to R^m)2012-01-21
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    @tomerga Do you know that every sequence in $[0,1]$ has a convergent subsequence? Try using it to prove that the inverse of $f$ would have to be continuous.2012-01-21
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    A little off topic: I was reviewing your question and found that, it surfaced here merely because of your short question. Try to keep your question self-contained, by adding the question and not referring to the title! And, Upvote and/or choose a top answer by clicking on the tick mark by the side of every answer. Express your gratitude to the answerers at the forum2012-01-21
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    I really don't see how to solve this problem using only the methods of multivariable calculus. You learn almost nothing about continuous functions in this course...only about differentiable ones. On the other hand, a little general topology makes this problem quite easy: as others have said, the key is that since $[0,1]$ is compact, the hypotheses force $f$ to be a homeomorphism, and it is easy to see that the interval is not homeomorphic to the square. I wonder what your instructor has in mind here? (I doubt it's the implicit function theorem for nondifferentiable functions...)2012-01-21

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