Proving a Set is ClosedDate: 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/ |
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