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Kate
Kate
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Is there a continuous function on $\mathbb$f:\mathbb{R}^2$ \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 on $\mathbb{R}^2$$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 . . .

Is there a continuous function on $\mathbb{R}^2$ 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 on $\mathbb{R}^2$ that has uncountably infinite turning points?"

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?

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 . . .

Source Link
Kate
Kate
  • 65
  • 1
  • 5

Is there a continuous function on $\mathbb{R}^2$ 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 on $\mathbb{R}^2$ that has uncountably infinite turning points?"

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?