Open vs. connected subsets

[vvaann]

Let f be a real-valued function defined and continuous on the set of real numbers R. Which of the following must be true of the set S = {f(c): 0 < c < 1} ?

I. S is a connected subset of R.
II. S is an open subset of R.
III. S is a bounded subset of R.

(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III

Answer key:

SPOILER: C

I'm confused by the answer key suggesting that S is a connected subset of R.
As {c: 0 < c < 1} is an open subset of R and f(c) is continuous, I think {f(c): 0 < c < 1} should be an open subset of R also. Isn't it??

[Dragonfinity]

The image of a connected subset under a continuous map is connected. The set (0,1) is connected, so its image S is connected. However, the image of an open set is not necessarily open. For example, if f is any constant function, then S is merely one point, which is a closed set.

[vvaann]

It's clear now. Thanks again, Dragonfinity.

[lmtuan]

According to Dragonfinity:
* S is a connected subset of R.
* S is not necessary open.
( I agree with Dragonfinity).
Since f is a real-valued function defined and continuous on the set of real numbers R, f is continuous on [0,1]. Thus, f([0.1]) is bounded. In other hand, we have S is a subset of f([0,1]). Hence, S is a bounded subset of R.
Answer: (C)

[kostya86]

Wait, but here we have a continuous function defined on an open interval (0,1). Thus, it does not necessarily follow that its image is a compact set, and thus the function may not attain a supremum on that set. So how can you say that the function is bounded there? I mean take function like 1/x defined on (0,1)...

[adviolin]

Kostya, you're missing the fact that the function is defined for all real numbers, so your example doesn't apply. Since it's defined for all numbers, it's defined on [0,1], thus f([0,1]) is compact, and thus bounded. Since f((0,1)) is a subset of this bounded set, it is bounded as well.

[lime]

Hi! That is really buffling question for me as well, but I cannot understand another variant:

Originally Posted by lmtuan

Since f is a real-valued function defined and continuous on the set of real numbers R, f is continuous on [0,1]. Thus, f([0.1]) is bounded.

*by the way the question was about (0,1) not [0,1].

WHY? If f=tan(Pi/2*(x-0.5)) so in that case we have (0,1)->(-inf, +inf) and it is not bounded? Where am I wrong?

[adviolin]

Lime,

You are right that a function continuous only on (0,1) need not be bounded, but that is not what this problem is asking.

This particular problem requires that f be continuous on all of R. In other words, it must be continuous EVERYWHERE! Your function does not satisfy this requirement, so the result is not true for your function.

Instead take a function that is DEFINED and CONTINUOUS at every number. Then the result is true by the several proofs given in this thread.

[lmtuan]

Originally Posted by lime

Hi! That is really buffling question for me as well, but I cannot understand another variant:

*by the way the question was about (0,1) not [0,1].

WHY? If f=tan(Pi/2*(x-0.5)) so in that case we have (0,1)->(-inf, +inf) and it is not bounded? Where am I wrong?

Your function is not defined at all real numbers.
I used [0,1] since (0,1) is a subset of [0,1]. ([0,1] is very nice, it is closed, bounded, compact, connected.) Since f is continuous in R, f also is continuous in [0,1].....

[tuanle]

Originally Posted by lmtuan

Your function is not defined at all real numbers.
I used [0,1] since (0,1) is a subset of [0,1]. ([0,1] is very nice, it is closed, bounded, compact, connected.) Since f is continuous in R, f also is continuous in [0,1].....

It looks like something about Real Analysis.