# Continuity & inverse image of closed sets being closed

• Mar 8th 2009, 11:07 AM
Last_Singularity
Continuity & inverse image of closed sets being closed
Question: Let $f$ be a function defined on a closed domain $D$. Show that $f$ is continuous if and only if the inverse image of every closed set is a closed set.

One approach is to solve it using the fact that the complement of open sets are closed. However, I am wondering how I can prove it using the definition of a closed set using limit points?

I know that a set is closed if it contains all of its limit points. How could I use this definition to prove the above question? In other words, how do I show that if a set in the range holds all of its limit points, mapping it backwards to the domain will create a set also containing all of its limit points?

Thanks a lot!
• Mar 8th 2009, 12:08 PM
Moo
Hello,

Here is something I did : http://www.mathhelpforum.com/math-he...790-post2.html , using neighbourhoods (continuous => ...)
The key for that is : a set is open if it is a neighbourhood of any of its point.

I'm not used to limit points... so I hope it will sort of help you !
• Mar 8th 2009, 12:32 PM
Plato
Quote:

Originally Posted by Last_Singularity
Question: Let $f$ be a function defined on a closed domain $D$. Show that $f$ is continuous if and only if the inverse image of every closed set is a closed set.
I know that a set is closed if it contains all of its limit points.

What sort of space are you working with? Is a metric space?
If it is a general top-space, if so what properties do it have?
To use the limit point approach you need to consider that.
• Mar 9th 2009, 03:29 PM
Last_Singularity
Quote:

Originally Posted by Plato
What sort of space are you working with? Is a metric space?
If it is a general top-space, if so what properties do it have?
To use the limit point approach you need to consider that.

This is real analysis, so $f: X \rightarrow Y$ where $X,Y \subseteq \mathbb{R}$
• Mar 9th 2009, 04:45 PM
Plato
Quote:

Originally Posted by Last_Singularity
This is real analysis, so $f: X \rightarrow Y$ where $X,Y \subseteq \mathbb{R}$

Point number 1. Real analysis is about more than $\mathbb{R}$ so why would you expect us to know in what setting your problem is casted?

Now to your question, for continuous functions we have the following.
The inverse image of an open set is an open set.
The complement of the inverse image is the inverse of the complement of the image.
The complement of a closed set is open.
If $M$ is closed then $M^c$ is open.
$f^{ - 1} \left( {M^c } \right) = \left( {f^{ - 1} (M)} \right)^c$
Can you finish?
• Mar 9th 2009, 05:07 PM
Last_Singularity
Yes, thanks a lot!

And sorry about not specifying that it was in a real analysis context; I was not aware that such concepts can be generalized to metric spaces and other areas of mathematics...