Why is a finite sub-cover necessary for proving continuity implies boundedness?

In summary, the problem statement is to prove that if a function is continuous on a closed interval, then it is bounded. The relevant information is that the previous exercise proved a similar result for a function that is continuous at a point, and suggests using the Heine-Borel theorem. The proof involves choosing a finite number of points and their corresponding bounds to cover the closed interval, rather than choosing a maximum bound from an infinite collection of bounds, as the function could potentially keep increasing the bound as we move along the interval. The reason for this is that a half-open interval, which is not compact, does not have a finite sub-cover, as shown by a counterexample.
  • #1
Oats
11
1
1. The problem statement:
Let ##f:[a, b] \rightarrow \mathbb{R}##. Prove that if ##f## is continuous, then ##f## is bounded.

2. Relevant Information
This is the previous exercise.
Let ##A \subseteq \mathbb{R}## , let ##f: A \rightarrow \mathbb{R}##, and let ##c \in A##. Prove that if ##f## is continuous at ##c##, then there is some ##\delta > 0## such that ##f|A \cap (c - \delta, c + \delta)## is bounded.
I have already proved this result, and the book states to use it to prove the next exercise. It also hints to use the Heine-Borel theorem.

The Attempt at a Solution

:[/B]
Since ##f## is continuous, for each ##c \in [a, b]##, ##f## is continuous at ##c##. By the previous exercise, for each ##c \in [a, b]##, there is ##\delta_c > 0## such that ##f|A \cap (c - \delta, c + \delta)## is bounded, say by ##K_c##. Since, for each ##c \in [a, b]##, ##c \in (c - \delta_c, c + \delta_c)##, we have that the collection ##\{(i - \delta_i, i + \delta_i)\}_{i \in [a, b]}## forms an open cover of ##[a, b]##. By the Heine-Borel theorem, this collection has a finite subcover. That is, there exists ##n \in \mathbb{N}## and ##q_1, \ldots, q_n \in [a, b]## for which ##(q_1 - \delta_{q_1}, q_1 + \delta_{q_1}), \ldots, (q_n - \delta_{q_n}, q_n + \delta_{q_n})## form an open cover for ##[a, b]##, and each is bounded by ##K_{q_1}, \ldots, K_{q_n}## respectively. Now take ##K = \text{max}\{K_{q_1}, \ldots, K_{q_n}\}##. Let ##x \in [a, b]##. Then there is ##q_h##, for ##h \in \{1, \ldots, n\}##, for which ##x \in (q_h - \delta_{q_h}, q_h + \delta_{q_h})##, so that ##|f(x)| \leq K_{q_h} \leq K##. Hence, ##f## is bounded by ##K##.

I feel quite confident in most of the proof, but at the beginning I was feeling a little iffy on exactly why we need a reduction on the cover of ##[a, b]## to a finite one. I immediately sensed that that's what they where after with the hint to use the Heine-Borel theorem, but the actual necessity for the reduction itself is what irked me. I was wondering why we couldn't simply let ##K = \text{max}\{K_c\}_{c \in [a, b]}##. Since then, I began to question that reasoning with the following response, of which I would greatly appreciate feedback in determining if I am right: The reason we cannot simply choose the bound from an infinite collection of bounds, is that the bounds may be increasing. Sure, the function may be bounded around arbitrarily small neighborhoods around each point, but there are unaccountably many of these neighborhoods, and there is nothing stopping the possibility that the function simply keeps continually increasing that bound as ##c## increases in ##[a, b]##. However, with the finitely many guaranteed by the Heine-Borel theorem, there would have to be a largest.
 
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  • #2
Oats said:
However, with the finitely many guaranteed by the Heine-Borel theorem, there would have to be a largest.
Yes, that's the reason.

If you want to try to solidify your intuition further, try to prove the theorem for the function ##f:[a,b)\to \mathbb R##, ie where the domain is a half-open interval. It can't be done. A counterexample is the function ##f(x)=\frac1{b-x}##. It increases without limit as ##x\to b##. So the ##K_c##s will have no upper bound. The reason the proof doesn't work in this case is that a half-open interval is not compact and so not every cover has a finite sub-cover. In particular, the cover defined in the proof will have no finite sub-cover.
 

Related to Why is a finite sub-cover necessary for proving continuity implies boundedness?

1. What does "continuity implies bounded" mean?

"Continuity implies bounded" is a mathematical concept that states if a function is continuous on a closed interval, then it is also bounded on that interval. This means that the function's values do not become infinitely large or small on that interval.

2. How is continuity related to boundedness?

Continuity and boundedness are two important properties of a function. Continuity refers to the smoothness of a function, while boundedness refers to the range of values that a function can take on. In the case of "continuity implies bounded," it means that a continuous function cannot have values that go to infinity on a closed interval.

3. Can a function be continuous but not bounded?

Yes, it is possible for a function to be continuous but not bounded. For example, the function f(x) = x on the interval [0, ∞) is continuous but not bounded. This is because as x approaches infinity, f(x) also approaches infinity.

4. Is "continuity implies bounded" a two-way implication?

No, "continuity implies bounded" is not a two-way implication. This means that while a continuous function is always bounded, a bounded function may not necessarily be continuous. For example, the function f(x) = 1/x on the interval (0, 1] is bounded but not continuous at x = 0.

5. What is the significance of "continuity implies bounded" in mathematics?

The concept of "continuity implies bounded" is important in various areas of mathematics, including calculus, real analysis, and differential equations. It allows us to make conclusions about a function's behavior without knowing its exact values and helps us understand the relationship between continuity and boundedness.

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