If f is continuous on [a,b], then f is bounded on [a,b].

In summary, the conversation centers around a user seeking help with a theorem in their Analysis II course. The theorem involves proving that a set S is equal to the interval [a,b] under certain conditions. The users discuss various approaches and techniques, such as using the fact that a continuous function maps compact sets to compact sets and invoking the intermediate value theorem. The user also mentions being a math major and requesting that explanations not be too arcane.
  • #1
Esran
73
0
Dear friends,

I just joined the forums, and I'm looking forward to being a part of this online community. This semester, I signed up for Analysis II. I'm a math major, so I should be able to understand pretty much everything you say (hopefully). However, I'd really appreciate it if you try not to be too arcane in your explanations.

So, here's the theorem I'm stuck on.

Question: Prove...

Suppose f is continuous on [a,b] and S is the set such that x is in S if and only if x is in [a,b] and, (1) x = a or (2) f is bounded on the subinterval [a,x]. Then S is [a,b].

This is what I have so far.

Attempted solution:

Let S = {x in [a,b] | x = a or f is bounded on [a,x]}
Thus, by definition, S is a subinterval of [a,b].
Because of this fact, S is bounded.
Because S is bounded, S has a least upper bound, call it p.
It follows that p > a must be true.
Because b is an upper bound of S, p <= b.

Now, we employ an indirect argument.
Thus, assume p < b.
By definition, b is the LUB of set [a,b].


Other questions, and relevant background:

From here on out, I'm confused about where to go and what to do. Does anyone have any pointers or suggestions? What approach do I need to take to complete this proof? The only thing I can think of is that it might have to do with f being continuous on an interval implying that f is bounded on that interval. However, I'm not sure how to prove this implication either.

Thank you in advance for your help.
 
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  • #2
It seems like you need to show there's an element of the set other than a. I mean, it's pretty obvious, but it's kind of needed, I think.

Of course, I would personally just say that a continuous function maps compact sets to compact sets, then apply Heine-Borel. But that's a copout.
 
  • #3
Hmmm...true, but remember. Ultimately, I'm trying to show that S is [a,b]. The continuous function bit is a part of the problem I think, though it might not be.
 
  • #4
Well, if it's discontinuous it could be that it's not bounded on any such interval [a,x] (take f(x) = a for x = a and f(x) = 1/(x - a), for x > a.
 
  • #5
It seems to me you can do this much more efficiently if you merely use the fact that f is continuous on [a,b]. I know you ask this later in your OP, but you haven't clearly used it anywhere in your outline. Remember that f continuous on [a,b] means f achieves a max/min on [a,b].
 
  • #6
Yeah, it seems like you're going to have to use some form of the compact sets being mapped to compact sets thing. If you want to show there's an x =/= a in the set, I think you're going to have to invoke the intermediate value theorem, which is a result of continuous functions taking on a maximum and minimum value on a closed interval, which is proved with the stated theorem and Heine-Borel.
 

FAQ: If f is continuous on [a,b], then f is bounded on [a,b].

What does it mean for a function to be continuous?

Continuous functions are those that have no sudden jumps or holes in their graphs. This means that the function can be drawn without lifting the pencil from the paper and there are no breaks or gaps in the graph.

How do you determine if a function is bounded?

To determine if a function is bounded, we need to look at its range or output values. If the function has a maximum and minimum value within a given interval, then it is considered bounded. In other words, the function does not grow infinitely large or small within the interval.

Why is it important for a function to be bounded on a given interval?

If a function is bounded on a given interval, then it means that the function does not have extreme or erratic behavior within that interval. This makes it easier to analyze and work with the function, as we can make predictions and solve problems more confidently knowing that the function will not have any unexpected behavior.

How does continuity relate to boundedness?

Continuity is a necessary condition for boundedness. This means that if a function is continuous on a given interval, then it is also bounded on that interval. This is because a continuous function does not have any sudden jumps or breaks in its graph, which would cause it to be unbounded.

Can a function be continuous but not bounded?

Yes, it is possible for a function to be continuous but not bounded. This can happen if the function has an asymptote or if it oscillates infinitely within a given interval. However, if the function is both continuous and defined on a closed interval, then it must be bounded.

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