Suppose f : D R is continuous with D compact. Prove that {x : 0 f(x) 1} is compact.

Since D is compact, it is closed and bounded. Thus, {x : 0 f(x) 1} is bounded.

Let x0 {x : 0 f(x) 1}.

Now, since f is continuous, for 0 = min (f(x0), 1 f(x0)), there is a for which x (x0 , x0 + ) D implies that |f(x) f(x0)| < epsion .

Since this is true of all x in this interval, define n = /n.

We can then construct {xn} such that xn (x0 n, x0 + n) for all n.

Thus, the sequence xn x0 (since n 0) is a sequence of members of D. Thus, x0 is a point of closure.

Thus, {x : 0 f(x) 1} is closed.

Is this proof correct?