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Intuitively, a function $\mathbb{R}\rightarrow\mathbb{R}$ is continuous if you can draw its graph without taking the pen off the page. This suggests the following theorem:

A map $f:X \rightarrow Y$ is continuous if and only if $f$ is connected in the product topology $X \times Y$.

Is this true? And if not, can anyone think of an additional premise or two that would make it true?

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    The example of @t.b. can be thought of as the map $\mathrm{Arg} : \mathbb{C}\setminus \{ 0 \} \to\mathbb{R}$ which associates to a complex number the _principal value_ of the argument of that number. Its graph is a piece of a [helicoid](http://en.wikipedia.org/wiki/Helicoid). In programming, that map is often called [`atan2`](http://en.wikipedia.org/wiki/Atan2) (link contains illustration).2014-08-06

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It isn't true in general. An obvious variant of the Topologist's sine curve provides an example of a function $f:\Bbb R\rightarrow \Bbb R$ whose graph is connected but fails to be continuous (at $x=0$).

However, this article shows that "it is correct to conclude that continuous real functions over $\Bbb R$ are those functions over $\Bbb R$ whose graphs, in the plane $\Bbb R^2$, are both closed and connected".

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    The following result seems to be related: Theorem 8.2 in van Rooij-Schikhof, [p.52](http://books.google.com/books?id=Cqk5AAAAIAAJ&pg=PA52) For a function $f \colon [a,b]\to\mathbb R$ the following are equivalent: $f$ is continuous. $\Leftrightarrow$ $\Gamma_f$ is compact. $\Leftrightarrow$ $\Gamma_f$ is arcwise connected. (Here $\Gamma_f$ denotes the graph of $f$.)2012-08-18