A question involving connectedness and constant functions

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Sorry, this is my 3rd time for asking for help in a week and a half. I just have completely lost any intuition I had for math with all the time constraits I have.

Homework Statement

a.)Prove that a continuous function, f:M->R, all of whose values are integers, is constant provided that M is connected.
b.)What if all the values are irrational?

Homework Equations

M is connected if it has no separation. A separation is defined as 2 nonempty open sets A and B such that A U B=M and A \capB= the empty set.

Ny interpretation for f is constant is that it has only one element in it's image. f(x)=c for all x.

The Attempt at a Solution

I proved a.) using contradiction. I assumed M was connected and f was not constant. I let A=the set of all x in M such that f(x)=c and B=the set of all x in M such that f(x) didn't equal c. I proved that both A and be were open and came to the conclusion that there was a separation. I can't use the same logic for part b.) as proving A was open required that the distance between each integer be a discrete value.

I'm assuming the answer for b.) is no but I don't know how to prove it.

 

[micromass]

a seems fine, even though it's not the standard way of proving it.

I'll give you a hint for b. Assume that a and b are in the image of f. Take a rational number q between a and b. You'll need to show that f^{-1}(]-\infty,q[) and f^{-1}([q,+\infty[) is a separation.

The standard way to prove both a and b, is to actually show that if X is connected, then f(X) is connected. Thus connectedness is preserved under continuous images. This makes it easier to show a and b...

 

[raw]

a seems fine, even though it's not the standard way of proving it.

I'll give you a hint for b. Assume that a and b are in the image of f. Take a rational number q between a and b. You'll need to show that f^{-1}(]-\infty,q[) and f^{-1}([q,+\infty[) is a separation.

The standard way to prove both a and b, is to actually show that if X is connected, then f(X) is connected. Thus connectedness is preserved under continuous images. This makes it easier to show a and b...

So I proved that f(X) is connected if X is connected. The first bit of advice confused me a little when I went to do it. Do you mean f^{-1} of the interval from q to infinity as only a and b are in the image.

 

[micromass]

Yes, I know that f^{-1}(]-\infty,q])=f^{-1}(a). But the advantage is that ]-\infty,q[ is an open set. Thus this shows that f^{-1}(a) is an open set. Similarly, f^{-1}(b) is an open sets. So you've found two disjoint open sets.

 

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Oh ok, that's what I thought and I just finished it. Thank you so much.