Continuity & inverse image of closed sets being closed

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March 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!
March 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 !
March 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.
March 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 $X,Y \subseteq \mathbb{R}$
March 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?
March 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...