The extreme values theorem says continuous functions achieve their maximum and minimum provided they are defined on a sequentially compact set.
Lemma 6.3.1 Let K ⊆ F be sequentially compact and let f : K → ℝ be continuous. Then f is bounded. That is there exist numbers, m and M such that for all x ∈

Proof: Suppose f is not bounded above. Then there exists
Example 6.3.2 Let f
Clearly, f is not bounded. Does this violate the conclusion of the above lemma? It does not because the end points of the interval involved are not in the interval. The same function defined on [.000001,1) would have been bounded although in this case the boundedness of the function would not follow from the above lemma because it fails to include the right endpoint.
The next theorem is known as the max min theorem or extreme value theorem.
Theorem 6.3.3 Let K ⊆ F be sequentially compact and let f : K → ℝ be continuous. Then f achieves its maximum and its minimum on K. This means there exist, x_{1},x_{2} ∈ K such that for all x ∈ K,

Proof: By Lemma 6.3.1 f
In fact a continuous function takes compact sets to compact sets. This is another of those big theorems which tends to hold whenever it makes sense. Therefore, I will be vague about the domain and range of the function f.
Proof: Suppose C is an open cover of f

Thus