Prove or disprove : if $f$ define on $A \subset \mathbb R^n$ and its graph is connected, then $f$ is continuous.

Question

I don't think this is true, I know that if set $A$ is path connected then $A$ is connected. However, $A$ is connected doesn't guarantee that it's path connected, thus we can't be sure that $f$ is continuous. I tried to find an counter example, but I can't find any.

You are right. For instance, Baire 1 functions $f:\mathbb R\to\mathbb R$ with the intermediate value property have connected graphs. Any derivative is an example of such an $f$. But there are discontinuous derivatives. — Andrés E. Caicedo, Mar 07 '14 at 17:21
The standard counterexample is probably the topologist's sine curve. $f(x) = \sin \frac1x$ for $x\neq 0$, $f(0) = 0$. — Daniel Fischer, Mar 07 '14 at 17:22

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

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What about $\sin {1 \over x},$ with value $0$ at $x=0$?

That function is not continuous (it doesn't have a limit as $x$ approaches $0$), its graph is connected but is not path connected. A proof can be found as an answer to this question.

It is a function to keep in mind when trying to find counter examples. For more information you can search for topologist's sine curve.

edited Apr 13 '17 at 12:20

Community

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answered Mar 07 '14 at 17:24

dafinguzman

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@dror I thought it would be enough with the hint. – dafinguzman Mar 07 '14 at 18:15
I reviewed an earlier version of this answer, so consider the first comment irrelevant now – user76568 Mar 07 '14 at 18:19

score 1 · Answer 2 · answered Mar 07 '14 at 17:30

In the space $R^3$ you can imagine a surface of a function f(x,y) that is connected but not continues (it may have wedges). See http://en.wikipedia.org/wiki/File:Riemann_surface_log.jpg (Riemann surface) in [0,2$\pi$) is connected but not continuous.

Prove or disprove : if $f$ define on $A \subset \mathbb R^n$ and its graph is connected, then $f$ is continuous.

2 Answers2