# Image of a f with a local minima at all points is countable.

• Terrell
In summary: No, it could be constant. It could be a step function (although there is a constraint on the steps - can you see what that is?). It could have isolated points.Must fff be increasing?Another idea: if the image is uncountable, then it must be uncountable in at least one of the intervals ##[n, n+1]##. That might help.Or, perhaps your original idea to use the density of ##\mathbb{Q}## might...work...somehow.I think this is it: For every ##x \in \Bbb{R}##, there is a corresponding ##\epsilon >0##. Then, there is no other number

## Homework Statement

Let ##f:\Bbb{R} \to \Bbb{R}## be a function such that ##f## has a local minimum for all ##x \in \Bbb{R}## (This means that for each ##x \in \Bbb{R}## there is an ##\epsilon \gt 0## where if ##\vert x-t\vert \lt \epsilon## then ##f(x) \leq f(t)##.). Then the image of ##f## is countable.

N/A

## The Attempt at a Solution

Let ##f:\Bbb{R} \to \Bbb{R}## be a function such that ##f## has a local minimum for all ##x \in \Bbb{R}##. Consider the arbitrary interval ##(r- \epsilon, r+ \epsilon)##. Since if ##q \in \Bbb{Q}## is in ##(r - \epsilon, r+ \epsilon)##, then ##(r - \delta, r + \delta) \subset (r - \epsilon, r + \epsilon)## such that ##\vert q-r\vert=\delta##. Hence we can define a mapping ##g:f(\Bbb{R}) \to \Bbb{Q}## as ##g(f(r))=\delta##. Note that we can assign a unique correspondence between elements in ##f(\Bbb{R})## and ##\Bbb{Q}## due to the density property; which states ##\forall x,y\in\Bbb{R} \exists q' \in \Bbb{Q}, x \lt q' \lt y##. Since ##\Bbb{Q}## is countable then so is ##f(\Bbb{R})##.

Terrell said:

## Homework Statement

Let ##f:\Bbb{R} \to \Bbb{R}## be a function such that ##f## has a local minimum for all ##x \in \Bbb{R}## (This means that for each ##x \in \Bbb{R}## there is an ##\epsilon \gt 0## where if ##\vert x-t\vert \lt \epsilon## then ##f(x) \leq f(t)##.). Then the image of ##f## is countable.

N/A

## The Attempt at a Solution

Let ##f:\Bbb{R} \to \Bbb{R}## be a function such that ##f## has a local minimum for all ##x \in \Bbb{R}##. Consider the arbitrary interval ##(r- \epsilon, r+ \epsilon)##. Since if ##q \in \Bbb{Q}## is in ##(r - \epsilon, r+ \epsilon)##, then ##(r - \delta, r + \delta) \subset (r - \epsilon, r + \epsilon)## such that ##\vert q-r\vert=\delta##. Hence we can define a mapping ##g:f(\Bbb{R}) \to \Bbb{Q}## as ##g(f(r))=\delta##. Note that we can assign a unique correspondence between elements in ##f(\Bbb{R})## and ##\Bbb{Q}## due to the density property; which states ##\forall x,y\in\Bbb{R} \exists q' \in \Bbb{Q}, x \lt q' \lt y##. Since ##\Bbb{Q}## is countable then so is ##f(\Bbb{R})##.

I don't see how your proof works. To take an example. Let ##r = 0## and ##\epsilon = 1##.

We can find ##q = 1/2##, say, and then ##g(f(0)) = q = 1/2##.

But, we could also choose ##q = 1/4## here.

And, if we take ##r = 0## and ##\epsilon = 0.1##, we would need a different ##q## again.

Terrell
PeroK said:
But, we could also choose q=1/4q=1/4q = 1/4 here.
Yeah I know. However, it is not what I meant. I am having difficulty writing it precisely. But, I think I've found another way. For each ##x \in (r-\epsilon, r+\epsilon)##, find an interval within ##(r-\epsilon, r+\epsilon)## with rational endpoints containing ##x## and assign either one of the rational endpoints to ##x##. Would this work?

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Terrell said:
Yeah I know. However, it is not what I meant. I am having difficulty writing it, precisely. But, I think I've found another way. For each ##x \in (r-\epsilon, r+\epsilon)##, find an interval within ##(r-\epsilon, r+\epsilon)## with rational endpoints containing ##x## and assign either one of the rational endpoints to ##x##. Would this work?

I don't see how. I don't know this problem, but here are my thoughts:

Why isn't ##f## simply the constant function?

How does ##f## change from one value to another?

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PeroK said:
Why isn't fff simply the constant function?
I think the author of the problem wants us to show for any such function ##f## and not just for a constant function.
PeroK said:
How does fff change from one value to another?
It wasn't specified in the book. Although, it was also hinted that we should avoid assuming additional properties like differentiability and continuity or anything like that.

He(Steven Krantz) also hinted that he had a 10 page solution when was a student. While a one line solution was provided by Michael Spivak.

Terrell said:
assign either one of the rational endpoints to x
I meant, assigned to ##f(x)## in particular.

Terrell said:
I think the author of the problem wants us to show for any such function ##f## and not just for a constant function.

It wasn't specified in the book. Although, it was also hinted that we should avoid assuming additional properties like differentiability and continuity or anything like that.

He(Steven Krantz) also hinted that he had a 10 page solution when was a student. While a one line solution was provided by Michael Spivak.

I didn't expect to be so comprehensively misunderstood.

Well, what sort of function do you think that ##f## could be? What would an ##f## with those properties look like?

PeroK said:
I didn't expect to be so comprehensively misunderstood.
Sorry for the misunderstanding.
PeroK said:
Well, what sort of function do you think that fff could be? What would an fff with those properties look like?
The way I see it, graphically, ##f## would be an infinite number of dots in ##\Bbb{R}^2##; I think it's called a piece-wise function. I think that ##f## cannot be continuous. Is this a correct interpretation to begin with? Thanks!

Terrell said:
graphically, fff would be an infinite number of dots in R2
Moreover, the dots must move upward of ##\Bbb{R}^2## to infinity. I think, this must be the case because for every interval ##(r-\epsilon, r+\epsilon)##, ##f(r) \leq f(x')## such that ##x' \in (r-\epsilon, r+\epsilon)##.

Terrell said:
Sorry for the misunderstanding.

The way I see it, graphically, ##f## would be an infinite number of dots in ##\Bbb{R}^2##; I think it's called a piece-wise function. I think that ##f## cannot be continuous. Is this a correct interpretation to begin with? Thanks!

No, it could be constant. It could be a step function (although there is a constraint on the steps - can you see what that is?). It could have isolated points.

Must ##f## be increasing?

Another idea: if the image is uncountable, then it must be uncountable in at least one of the intervals ##[n, n+1]##. That might help.

Or, perhaps your original idea to use the density of ##\mathbb{Q}## might help.

Terrell
Terrell said:
Moreover, the dots must move upward of ##\Bbb{R}^2## to infinity. I think, this must be the case because for every interval ##(r-\epsilon, r+\epsilon)##, ##f(r) \leq f(x')## such that ##x' \in (r-\epsilon, r+\epsilon)##.

This is not true. ##f## could be bounded.

PeroK said:
(although there is a constraint on the steps - can you see what that is?). It could have isolated points.
I don't see this, yet. Don't know if I will, I'll try.
PeroK said:
Must fff be increasing?
Nope.

PeroK said:
This is not true. fff could be bounded.
Only when ##f## is a constant?

Terrell said:
Only when ##f## is a constant?

You could have a step function with a finite number of steps. Or, an infinite number of steps, as long as the sequence of step values is bounded.

Terrell
PeroK said:
You could have a step function with a finite number of steps. Or, an infinite number of steps, as long as the sequence of step values is bounded.
Got it. Although, bounded-ness does not affect this problem. Correct?

PeroK said:
Another idea: if the image is uncountable, then it must be uncountable in at least one of the intervals [n,n+1][n,n+1][n, n+1]. That might help.
I haven't thought of this equivalent statement. This is nice.

Terrell said:
I haven't thought of this equivalent statement. This is nice.

Okay. I might have time to look for a solution myself today. But, if I don't, at least you have some more ideas.