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Intuition on why there can't be a continuous bijection between $(a,b)$ and $[c,d]$?

I'm not (necessarily) looking for a proof for this, I want to understand why does this happen, intuitively: if I add uncountably many points to $(a,b)$ taking the set $(a,b+1)$ I can find a continuous bijection, but if I only add the extremes, this is impossible, why?

Stefan Hamcke
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YoTengoUnLCD
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    The intuition, I'd say, is that some points have to be mapped to the endpoints, and then the map cannot be injective at these points. – Stefan Hamcke Sep 10 '15 at 16:49
  • Here is a similar question that already got answer:http://math.stackexchange.com/questions/42308/continuous-bijection-from-0-1-to-0-1. So the problem is possibly duplicated. – Terence Hang Sep 10 '15 at 16:54
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    One is compact and the other is not. That rules out any continuous bijection from the one that is compact to the one that is not, because a continuous image of a compact space is compact. ${}\qquad{}$ – Michael Hardy Sep 10 '15 at 17:21

2 Answers2

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Suppose such a function exists with $x\in (a,b)$ such that $f(x)=d$. Pick points $y,z\in(a,b)$ with $y<x$ and $z>x$. We know such points exist because $x$ is an interior point of $(a,b)$. Without loss of generality, suppose $f(y)<f(z)$. We know that both $f(y)$ and $f(z)$ are less than $d$. By the intermediate value theorem, there exists $u\in[y,x]$ such that $f(u)=f(z)$. This is a contradiction to the hypothesis that $f$ is injective, so $f$ cannot exist.

Plutoro
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Any continuous injective map on an interval must be strictly monotone. Let's assume that the map $f:(a, b) \to [a, b]$ is bijective and strictly increasing. Then there must be a point $c \in (a, b)$ that satisfies $f(c) = a$. But since $c > a$, all the points in the interval $(a, c)$ must be mapped to a value that is strictly less than $a$. This is impossible, since $a$ is already the smallest number in the range of $f$.

The case that $f$ is strictly decreasing works in a similar fashion by considering the interval $(c, b)$.

Dominik
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