Proving a Set is Closed

Date: 04/15/98 at 09:37:28
From: Okie
Subject: real analysis

Q: Let f be a continuous real valued function on Rn.
   Let S = {x in Rn : f(x) = 1}. Prove that S is a closed 
   subset of Rn .

I'm not sure how to go about doing this question, especially how to 
set it out.

Thank you for your help.

Date: 04/15/98 at 12:38:42
From: Doctor Kate
Subject: Re: real analysis

Well, usually to set out proofs, we want to look at the tools we can 
use. Here, we need a definition of 'closed' before we can prove 
anything about it. The definition I am familiar with is:

   A set F in Rn is said to be closed if, whenever {x_k} is a           
   sequence in F which converges to some x_0, x_0 must belong 
   to F.

Note: I am using the symbol _ to represent a subscript. So x_0 is "x 
sub zero" or "x naught".

Now, the only other information we have available is the definition of 
S and the fact that f is a continuous real valued function on Rn.  
Hmm, continuous. What do we know about continuous functions (it is 
probably of some importance since they put it in there)?

Well, one thing I know about continuous functions, and you probably do 
too, is that if f is continuous then if {x_k} converges to some x_0, 
{f(x_k)} converges to f(x_0).

Now, if you think about these things and put them together, you should 
have a proof. It will have bare bones like this:

Using the definition of closed...
Take a sequence in S that converges to some x_0. We want to show 
x_0 is in S. Well, assume it is not. Then f(x_0) is not equal to 1. 
But by continuity, {f(x_k)} is converging to f(x_0), where {x_k} are 
the elements in the sequence (each of which is in S). Is there a 
problem here? Can x_0 not be in S? What is the value of each f(x_k)? 
Can these values converge to f(x_0)?

Good luck.

-Doctor Kate,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/