Dear friends,(adsbygoogle = window.adsbygoogle || []).push({});

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.

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.

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].

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**