Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is the problem: I have a space $X=\{a,b\}$ with trivial topology (open sets are empty set and whole $X$) and continuous map $f\colon X\to Y$. ($Y$ is arbitrary space). Can I conclude that $f(a)=f(b)$ because original of some open set in $Y$ is $\{a,b\}$, and original of some closed set in $Y$ is $\{a,b\}$? My friend used it but I don't agree.

share|improve this question

3 Answers 3

Generally you cannot use that argument. For instance we might ask $Y=X$ and we also give $Y$ the trivial topology. Then the identity map is continuous but $f(a)\neq f(b)$.

In general, when the target space is in trivial topology, all functions are continuous.

share|improve this answer

No you cannot conclude this.

Consider the topology on $Y=\{a,b,c\}$ where $\{\varnothing,\{c\},\{a,b\},Y\}$ are open; and $f\colon X\to Y$ is the identity function.

share|improve this answer

You can conclude that $f(a)$ and $f(b)$ are topologically indistinguishable, i.e., every open set either contains both $f(a)$ and $f(b)$ or contains neither. In general, you cannot conclude $f(a) = f(b)$. However, if $Y$ satisfies the $\mathrm{T}_0$ separation axiom, then $f(a) = f(b)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.