How Can a Function from Real Numbers to a Discrete Space Be Continuous?

[Silviu]

Homework Statement

Find $f:R \to X$, f-continuous, where X is the discrete space.

Homework Equations

The Attempt at a Solution

f is continuous at p if for any $\epsilon > 0$ there is $\delta >0$ such that $d(f(x),f(p))<\epsilon$ for all x such that $d(x,p)<\delta$. Let $\epsilon = 1$. As X is the discrete space, only f(p) satisfies $d(f(x),f(p))<\epsilon$. So for a whole open interval (which becomes closed due to continuity) we have $f([p-\delta,p+\delta])=f(p)$. From here I can see that only constant functions would work here, but I am not sure how to continue. I was thinking to do the same reasoning at $f(p+\delta)$, and extend this over the whole R, but if the intervals get smaller and smaller this might not work. How should I continue?

 

[mfb]

What do you want to continue?
You can formally show that f(x)=f(0) by choosing a suitable interval if you like.

If $f(x) \neq f(0)$, then you can find at least one point of disconinuity in between.

 

[Silviu]

What do you want to continue?
You can formally show that f(x)=f(0) by choosing a suitable interval if you like.

How can I choose this interval? I don't know how $\delta$ depends on $\epsilon$. I just know it exists

 

[Ray Vickson]

How can I choose this interval? I don't know how $\delta$ depends on $\epsilon$. I just know it exists

For some continuous functions, $\delta$ depends on $\epsilon$, but for other continuous functions it does not.

 

[Silviu]

For some continuous functions, $\delta$ depends on $\epsilon$, but for other continuous functions it does not.

Yes, but in my case I need to consider the most general case. So How do I do?

 

[Ray Vickson]

Yes, but in my case I need to consider the most general case. So How do I do?

You need to think more about your problem; we are not allowed to solve it for you.

However, here is a hint: what do you mean by a "discrete space"?

 

[Silviu]

You need to think more about your problem; we are not allowed to solve it for you.

However, here is a hint: what do you mean by a "discrete space"?

That means that $d(x,y)$ is 0 if x=y and 1 otherwise. This is what I used to get a neighborhood of a point that contains just that point and thus an interval from R will be completely sent to that point. I did this. I don't know how to continue from here. I don't want a proof just a suggestion.

 

[PeroK]

That means that $d(x,y)$ is 0 if x=y and 1 otherwise. This is what I used to get a neighborhood of a point that contains just that point and thus an interval from R will be completely sent to that point. I did this. I don't know how to continue from here. I don't want a proof just a suggestion.

My first thought would be to show that a non-constant function is not continuous.