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How can we show, rigorously, that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous map then its graph and the inverse of the graph (i.e the inverse of the relation) is a nowhere dense set? I'm trying to show that the plane cannot be covered by a countable number of graphs of continuous maps and its inverses, so by showing the above the result follows from the Baire category theorem.

Thanks in advance

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The graph $\Gamma(f)$ of a continuous function $f$ is closed, so we just have to show it has empty interior. But of course if $\Gamma(f)$ had nonempty interior it could not pass the "vertical line test" and would not be a graph. For the inverse, just turn your head sideways.

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The "hard" part is showing that the graph of f, $ \Gamma(f) = \{(x,f(x)) \} \subset \mathbb{R} \times \mathbb{R}$ is closed. It's then clear that no point in this subset can be in the interior of $\Gamma(f)$ for the reason Nate stated above.

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Is this really any harder than the rest? Take a point $(x,y)$ in the complement of the graph. Since $y \neq f(x)$ and $f$ is continuous, there are $\epsilon$ and $\delta$ such that on the $\delta$-interval around $x$, the values of $f$ are at least $\epsilon$ away from $y$, which gives an open rectangle about $(x,y)$ in the complement of the graph. – Pete L. Clark Jan 9 '11 at 8:28

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