# proving a function is continuous with a given topology

• Jan 11th 2011, 06:33 PM
oblixps
proving a function is continuous with a given topology
let f: R -> R be injective where R is given by the cofinite topology. Show that f is continuous.

so i know that the cofinite topology consists of sets that are either empty or have finite complements. in the case of R, the open sets in the topology are intervals containing + and - infinite while leaving out finitely many terms. to prove that f is continuous i have to show that given any open set V in the codomain, the preimage f^-1 (V) in the domain is also open. i also know that in order for a set to be open in R, there must exist a ball centered at each point and contained within the set for every point in that set.

i can see intuitively that since f is injective, the cardinality of V is less than or equal that of the preimage of V. and since V is open the preimage of V should be open as well. however that was based on my intuition and probably does not constitute a formal proof. how would you show that the preimage of any V is open?
• Jan 11th 2011, 06:38 PM
Jose27
It's probably best to use the characterization of continuity in terms of closed sets, which are just finite sets in this case, and use the injectivity.
• Jan 12th 2011, 08:35 AM
oblixps
so far i've only learned the characterization of continuity in terms of open sets. is it posible to use that to derive the characterization in terms of closed sets?
• Jan 12th 2011, 09:09 AM
FernandoRevilla
Quote:

Originally Posted by oblixps
so far i've only learned the characterization of continuity in terms of open sets. is it posible to use that to derive the characterization in terms of closed sets?

Yes, it is. Suppose $\displaystyle f:X\rightarrow Y$ is continuous and let $\displaystyle F$ be a closed set of $\displaystyle Y$, then $\displaystyle F^c$ is open, and so $\displaystyle f^{-1}(F^c)=(f^{-1}(F))^c$ is open in $\displaystyle X$, therefore $\displaystyle f^{-1}(F)$ is closed.

Conversely, suppose that $\displaystyle F$ closed in $\displaystyle Y$ implies $\displaystyle f^{-1}(F)$ closed in $\displaystyle X$. Consider an open set $\displaystyle G$ of $\displaystyle Y$, then $\displaystyle G^c$ is closed in $\displaystyle Y$ and so, $\displaystyle f^{-1}(G^c)=(f^{-1}(G))^c$ is closed in $\displaystyle X$. Therefore $\displaystyle f^{-1}(G)$ is open in $\displaystyle X$ i.e. $\displaystyle f$ is continuous.

Fernando Revilla
• Jan 12th 2011, 03:07 PM
oblixps
thanks for the proof!

now to prove this i need to show that any preimage of a closed set V in R is closed as well. the closed sets in this case are finite sets and since f is injective, the preimage of f must contain the same number of elements or less than the closed sets in the codomain. so the preimages are finite as well and since a ball around any point in a finite set in R necessarily is not contained in the finite set, the preimages must be closed sets. therefore f is continuous.

would this be the correct reasoning?
• Jan 12th 2011, 05:48 PM
Drexel28
Quote:

Originally Posted by oblixps
thanks for the proof!

now to prove this i need to show that any preimage of a closed set V in R is closed as well. the closed sets in this case are finite sets and since f is injective, the preimage of f must contain the same number of elements or less than the closed sets in the codomain. so the preimages are finite as well and since a ball around any point in a finite set in R necessarily is not contained in the finite set, the preimages must be closed sets. therefore f is continuous.

would this be the correct reasoning?

Correct! Put more concisely since $\displaystyle f$ is injective we have that for any $\displaystyle C$ closed in $\displaystyle \mathbb{R}_C$ we have that $\displaystyle \#\left(f^{-1}\left(C\right)\right)\leqslant \#\left(C\right)<\infty$ and so $\displaystyle f^{-1}(C)$ is closed.