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Fix distinct $a,b \in \mathbb{R}^2$. In terms of cardinality (say, beth numbers), how many distinct continuous functions $f : [0,1] \rightarrow \mathbb{R}^2$ satisfying $f(0)=a, f(1)=b$ are there? (Presumably, the answer is independent of which $a,b$ are chosen.)

It seems clear that the answer is either $\beth_1$ or $\beth_2$.

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up vote 11 down vote accepted

There are only $\beth_1$ continuous functions from $[0,1]$ to $\Bbb R^2$, so there cannot be more than $\beth_1$, to see this note that every two continuous functions agreeing on a dense set are equal, as $[0,1]$ is separable there is a countable set which "decides" the values of the function. So there are no more than $(\beth_1)^{\aleph_0}=\beth_1$ continuous functions.

Clearly there are $\beth_1$, because take any $c\in\Bbb R^2$, there is a piecewise-linear path such that $f(0)=a,\ f(\frac12)=c,\ f(1)=b$. There are $\beth_1$ such $c$, therefore there are exactly $\beth_1$ many paths.

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I agree on the second paragraph, but I think the first needs some more explanation. – Axel Mar 30 '13 at 10:20

Clearly there are at least $\beth_1$ paths: for any $x \in (0,1)$, consider the path that sits in $a$ until time $x$ and then heads towards $b$ at constant speed. For the upper bound, note that a continuous function from $[0,1]$ to $\Bbb{R}^2$ is determined by its values at rational points, so there are at most $(\beth_1^2)^{\aleph_0}=\beth_1$ such functions.

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