It's intuitive to me to assume that for any continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ that is defined for all real numbers, it's curve plotted in $\mathbb{R}^2$ is homeomorphic to $\mathbb{R}$. But it then occured to me that the "arclength" between any two points on the curve of the Weierstrass function is probably infinite the same way it is on the Koch curve-and it doesn't make intuitive sense for a metric space (if it can even be classified as one) like this to be homeomorphic to the real line.
Is the Weierstrass function curve homeomorphic to the real line?
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general-topology
fractals
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0Yes - the projection $(x,y)\to x$ is a continuous bijection with a continuous inverse. Thus, the graph of a function $f:\mathbb R \to \mathbb R$ is homeomorhphic to $\mathbb R$. In general, the projection will not increase distance but it may decrease it substantially, as you've observed. – 2017-01-03
1 Answers
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They are indeed homeomorphic: the projection onto the $x$ axis is a homeomorphism from the graph of the Weierstrass function to the line, and indeed this is true for any continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$.
Note that arclength is not the only way to define a metric on this curve (and indeed, it doesn't define a metric, since it's always infinite between any two distinct points). But this doesn't actually matter. Just because there is some natural candidate for defining a metric on a certain space which doesn't actually do so, doesn't mean that there is no way to define a metric.