Proving f(0)=f(1) in the Sierpinski Topology with a Continuous Function

[Ted123]

Homework Statement

Let $X=\{0,1\}$ with the Sierpinski topology $\tau = \{ \emptyset , \{0\} ,\{0,1\} \}$.

Suppose $f:X\to \mathbb{R}$ is continuous.

Show $f(0)=f(1)$.

[Potentially useful observation: $\{f(0)\}$ is closed in $\mathbb{R}$.]

The Attempt at a Solution

$f:X\to\mathbb{R}$ is continuous $\iff$ for every open (closed) set $A\subseteq \mathbb{R},\;f^*(A)$ is open (closed) in $X$.

How to show f(0)=f(1)?

 

[Citan Uzuki]

Suppose the contrary, and find an open set whose preimage is {1}.

 

[Ted123]

Suppose the contrary, and find an open set whose preimage is {1}.

Does $f^*(\{0\}) = \{1\}$?

 

[Dick]

Does $f^*(\{0\}) = \{1\}$?

Who could say until you define what f is?? Follow Citan Uzuki's advice and define f(0)=a and f(1)=b where a and b are unequal real numbers. Is f continuous?

 

[Ted123]

Who could say until you define what f is?? Follow Citan Uzuki's advice and define f(0)=a and f(1)=b where a and b are unequal real numbers. Is f continuous?

So $\{a\} \subseteq \mathbb{R}$ and $\{b\} \subseteq \mathbb{R}$ are closed, and $f^*(\{a\})=\{0\} \subseteq X$ is closed in $X$ but $f^*(\{a\})=\{1\}$ is not closed in $X$ so $f$ is not continuous - contradiction; therefore $f(0)=f(1)$?

 

[Ted123]

So $\{a\} \subseteq \mathbb{R}$ and $\{b\} \subseteq \mathbb{R}$ are closed, and $f^*(\{a\})=\{0\} \subseteq X$ is closed in $X$ but $f^*(\{a\})=\{1\}$ is not closed in $X$ so $f$ is not continuous - contradiction; therefore $f(0)=f(1)$?

^ That should say $f^*(\{b\})=\{1\}$ but I can't find an edit button!

 

[Dick]

So $\{a\} \subseteq \mathbb{R}$ and $\{b\} \subseteq \mathbb{R}$ are closed, and $f^*(\{a\})=\{0\} \subseteq X$ is closed in $X$ but $f^*(\{a\})=\{1\}$ is not closed in $X$ so $f$ is not continuous - contradiction; therefore $f(0)=f(1)$?

Right. Now to finish up show that if a=b then f IS continuous.

 

[Citan Uzuki]

That approach will work, but you're doing it backwards. {0} is not closed in X, because its complement is {1}, which is not in the topology on X. So the conclusion that f is not continuous comes from the fact that f*({a}) = 0, not the fact that f*({b}) = 1.

 

[Dick]

^ That should say $f^*(\{b\})=\{1\}$ but I can't find an edit button!

Yeah, actually I didn't notice, but you right. My 'Edit' button disappeared for a while yesterday. Seems to be back now. At least for me. Citan Uzuki is also right. I didn't notice you'd swapped the open and closed either. Guess I should read more carefully.

 

[Ted123]

Yeah, actually I didn't notice, but you right. My 'Edit' button disappeared for a while yesterday. Seems to be back now. At least for me. Citan Uzuki is also right. I didn't notice you'd swapped the open and closed either. Guess I should read more carefully.

If we have a contradiction after supposing $a\neq b$ doesn't that mean we must have $a=b$?

 

[Dick]

If we have a contradiction after supposing $a\neq b$ doesn't that mean we must have $a=b$?

No! All you've shown is that if a not equal b is then f is not continuous. How can that be a proof that if a=b then f is continuous? There might not be ANY continuous functions.

 

[Ted123]

No! All you've shown is that if a not equal b is then f is not continuous. How can that be a proof that if a=b then f is continuous? There might not be ANY continuous functions.

But we're assuming all along that $f:X\to \mathbb{R}$ is continuous (as the question tells us this!)

 

[Dick]

But we're assuming all along that $f:X\to \mathbb{R}$ is continuous (as the question tells us this!)

Ok, this is really not my day. You are right. I thought you had to prove f continuous iff a=b. I saw the double arrow in your original post and fixated on it.