If f is contin. on [a,b], its range is a closed interval

[o_O]

I'm doing an old exam for my calculus class and the last question stumped me:

Question: If f is continuous on [a,b], show that the range of f is a closed interval [c,d]. (Hint: Be careful. You have to prove that the whole interval [c,d] is covered.

I'm really not sure how one would be able to approach this.
If f is continuous on [a,b], this implies $\displaystyle \lim_{x \to m} f(x) = f(m)$ for m $\displaystyle \in \left[a,b\right]$. That would mean I would have to somehow prove that f(m) $\displaystyle \in [c,d]$ ...

 

[pka]

How much topology do you know?
This is basically the theorem: The continuous image of a compact set is compact. In the reals that is equivalent to saying the continuous image of a closed and bounded set is a closed and bounded set.

Are you working with open covers that have finite subcovers?
If not work with sequence. Assume it is not bounded get a sequence that has no limit.

If you can tell us the exact context, what theorems and definitions you can use, then it is a lot easier to help. I don’t want to type up an argument that you cannot follow.

 

[o_O]

Sorry I haven't been formally introduced to any topology concepts. In fact, I'm only in a first-year calculus class looking over a mid-term over here. As you can tell from the context of the questions, we've only covered limits, continuity, and a bit of derivatives. So when I came across this question, it seemed to be a bit out of place.

Also, our prof wasn't referring to the min-max theorem here in the question before the one I posted in the PDF file in case you were wondering.[/url]

 

[pka]

Also, our prof wasn't referring to the min-max theorem

Well given your level of study, I am confused as to what the instructor expected!
If I were you, I would have said that the max/min theorem gives us the upper and lower bounds and the intermediate value theorem guarantees the image is an interval.

 

[o_O]

Min-max theorem according to our prof:
If f is a function continuous on the closed interval [a,b], then f attains its absolute maximum and its absolute minimum values on [a,b] (i.e. $\displaystyle \exists x_{1}$ and $\displaystyle x_{2}$ in [a,b] such that $\displaystyle \forall$x in [a,b] we have that $\displaystyle f(x_{1}) \leq f(x) \leq f(x_{2})$

OH. So by this theorem wouldn't f(x1) and f(x2) be our bounds for [c,d] and the intermediate value property tell us that there exists any number in [a,b] which would yield, through f, a number in the interval [c,d]. Somewhat correct?

I think the word 'prove' in the question intimidated me. Thanks a lot pka!

 

[pka]

the intermediate value property tell us that there exists any number in [a,b] which would yield, through f, a number in the interval [c,d].

BE CAREFUL. The intermediate value theorem says that: $\displaystyle f\left( {x_1 } \right) < t < f\left( {x_2 } \right)\quad \Rightarrow \quad \left( {\exists s \in \left[ {a,b} \right]} \right)\left[ {f(s) = t} \right]$.

 

[o_O]

Oh. So if for some values of s we get certain values of t which are in the interval [c,d], how can we prove that ALL values of s would satisfy the statement f(x1) < t < f(x2)?

 

[pka]

Oh. how can we prove that ALL values of s would satisfy the statement f(x1) < t < f(x2)?

The whole point is just that: $\displaystyle \left( {\forall t \in [c,d]} \right)\left[ {c = f\left( {x_1 } \right) < t < f\left( {x_2 } \right) = d} \right]\quad \Rightarrow \quad \left( {\exists s \in \left[ {a,b} \right]} \right)\left[ {f(s) = t} \right]$

 

[o_O]

Sorry for my misunderstanding but isn't that just showing that there exists some specific value of s such that f(s) = t. That doesn't cover all values of s in [a,b] ...

 

[pka]

Sorry for my misunderstanding but isn't that just showing that there exists some specific value of s such that f(s) = t. That doesn't cover all values of s in [a,b] ...

For all means for all. You just need to sit down and think about it.
$\displaystyle \vec{f\;} [a,b] \subseteq \left[ {c = f\left( {x_1 } \right),f\left( {x_2 } \right) = d} \right].$

 

[o_O]

Oh right, in the min-max theorem, it stated that ALL x in [a,b] will make f(x) be bounded by c and d. Thanks a lot pka!