# Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points?

I was thinking about the statement "if f is continuous on the interval I, there is not necessarily an interval J in I on which f is monotone." and this led me to the question "does there exist a continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ that has uncountably infinite turning points?" when I say turning point I'm talking about a point (x,f(x)) s.t there exists an open ball around that point where f(x) is either the highest or lowest value within that ball.

eg. $f(x)=sin(x)$ has countably infinite turning points as opposed to $f(x)=x^2$ which has one.

I cant think of a reason that convinces me that its impossible yet I can conceptualize a function that does this. Is it impossible? or does there exist such a function? I certainly get the impression this is impossible . . .

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You need to be more specific. The (graph of the) real-valued function f(x,y) = x - x^2 has uncountably many points (x,y) with a partial derivative of 0 and second partial negative. It is likely there are 2-D versions of Brownian motion which might come closer to what you actually intend to visualize. Gerhard "Ask Me About System Design" Paseman, 2011.03.29 – Gerhard Paseman Mar 30 '11 at 5:55
Please use the "edit" link below the question, and describe the definition of "turning point" that you are using. – S. Carnahan Mar 30 '11 at 11:11
Sorry for being so late to edit and for being vague, I think the definition of turning point I use above makes it impossible. – Kate Mar 30 '11 at 12:20
Still confusing. You ask about functions on ${\bf R}^2$ but your examples are of functions on $\bf R$ - only the graph is in ${\bf R}^2$. So what do you mean? – Gerry Myerson Mar 30 '11 at 12:24
Ah - you answered my question while I was typing it in. Thanks. – Gerry Myerson Mar 30 '11 at 12:25