vvaann Posted February 16, 2005 Share Posted February 16, 2005 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©: 0 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 © I and III only (D) II and III only (E) I, II, and III Answer key: C I'm confused by the answer key suggesting that S is a connected subset of R. As {c: 0 Quote Link to comment Share on other sites More sharing options...
Dragonfinity Posted February 17, 2005 Share Posted February 17, 2005 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. Quote Link to comment Share on other sites More sharing options...
vvaann Posted February 17, 2005 Author Share Posted February 17, 2005 It's clear now. Thanks again, Dragonfinity. Quote Link to comment Share on other sites More sharing options...
lmtuan Posted March 3, 2005 Share Posted March 3, 2005 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: © Quote Link to comment Share on other sites More sharing options...
kostya86 Posted March 4, 2007 Share Posted March 4, 2007 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)... Quote Link to comment Share on other sites More sharing options...
adviolin Posted April 6, 2007 Share Posted April 6, 2007 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. Quote Link to comment Share on other sites More sharing options...
lime Posted November 2, 2007 Share Posted November 2, 2007 Hi! That is really buffling question for me as well, but I cannot understand another variant: 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? :( Quote Link to comment Share on other sites More sharing options...
adviolin Posted November 8, 2007 Share Posted November 8, 2007 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. Quote Link to comment Share on other sites More sharing options...
lmtuan Posted December 13, 2007 Share Posted December 13, 2007 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]..... 1 Quote Link to comment Share on other sites More sharing options...
tuanle Posted December 13, 2007 Share Posted December 13, 2007 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. Quote Link to comment Share on other sites More sharing options...
lmtuan Posted December 13, 2007 Share Posted December 13, 2007 yup. You can see it in the Battle's book about Real Analysis. Quote Link to comment Share on other sites More sharing options...
anushrimali Posted November 6, 2008 Share Posted November 6, 2008 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? :( lime, (0,1) doesnt map to (-inf,+inf) with that function. the range is (-inf,1). Substitute x=1 and you will get f= tan (pi/2*1/2)= tan (pi/4)=1. Quote Link to comment Share on other sites More sharing options...
lime Posted November 6, 2008 Share Posted November 6, 2008 lime, (0,1) doesnt map to (-inf,+inf) with that function. the range is (-inf,1). Substitute x=1 and you will get f= tan (pi/2*1/2)= tan (pi/4)=1. My bad. I meant f(x)=(Pi)(x-1/2). Anyway, I understand where I was wrong. So never mind. Quote Link to comment Share on other sites More sharing options...
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