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Am I correct over statements below?

  1. The limsup and liminf of the sequence $n^2$ (meaning $1,4,9,16,\dots$) are equal. T

  2. Every bounded sequence has at most one convergent subsequence. F

  3. Are the following characteristic functions Riemann integrable on the interval $[0,1]$?

    • $\chi_{\left[0,\frac12\right]}$ yes
    • $\chi_{\Bbb Q}$ no
    • $\chi_C$, where $C$ is the Cantor set yes
    • $\chi_{\Bbb R-\Bbb Q}$ no
    • $\chi_{\left\{\frac1n:n\in\Bbb N\right\}}$ no
  4. No continuous function $f:\Bbb R\to\Bbb R$ can have a minimum value. (False)

  5. Let $I_1\supset I_2\supset I_3\supset\dots$ be a nested sequence of closed intervals in $\Bbb R$ whose lengths form a decreasing sequence converging to $0$. Choose points $a_n\in I_n$ for each $n$. Then the sequence $a_n$ converges, (I think it’s true)

  6. Consider a function $f:\Bbb R\to\Bbb R$. Which of the following statements are true?

    • If $f$ is continuous, then it maps every compact set onto a compact set? yes
    • If $f$ maps every compact set onto a compact set, then it is continuous. no
    • If $f$ is continuous, then it maps every connected set onto a connected set? yes
    • Is it true that if $f$ maps every connected set onto a connected set, then it is continuous. no
    • Is it true that if $f$ is continuous, then it maps every open set onto an open set? yes
    • If $f$ maps every open set onto an open set, then it is continuous. yes

(The original image from which this is copied is here.)

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Copying a page from a book is not the way questions are asked on this site! – Fabian Dec 16 '12 at 20:24
Sorry, there are questions I got partially correct on a online quiz. It didn't specify which ones are correct or wrong. So I type in word. Put it as a image and ask here. – user48601 Dec 16 '12 at 20:28
Please don’t vandalize the question. – Brian M. Scott Dec 16 '12 at 20:44
@Fabian: A close look at the image made it obvious that this was from an online exercise of some sort; you can even see where Enter was hit after the answers were typed in. It would have been more readable had Moriah copied it out, but it clearly isn’t a case of copying a page from a book. It’s a perfectly good question. – Brian M. Scott Dec 16 '12 at 20:46

1 Answer 1

up vote 0 down vote accepted

The last two parts of (6) are wrong. First, the constant function $f(x)=0$ is continuous, but the only open set that it maps to an open set is $\varnothing$. For the other example, define an equivalence relation $\sim$ on $\Bbb R$ by $x\sim y$ iff $x-y\in\Bbb Q$. For each $x\in\Bbb R$ the $\sim$-equivalence class of $x$ is $x+\Bbb Q=\{x+q:q\in\Bbb Q\}$, which is clearly dense in $\Bbb R$. Since each $\sim$-equivalence class is countable, there are $|\Bbb R|$ of them, so there is a bijection $\varphi$ from $\Bbb R/\sim=\{x+\Bbb Q:x\in\Bbb R\}$, the set of $\sim$-equivalence classes, to $\Bbb R$. Now define

$$f:\Bbb R\to\Bbb R:x\mapsto\varphi(x+\Bbb Q)\;.$$

Every open interval in $\Bbb R$ contains a member of each $\sim$-equivalence class, so $f$ maps each open interval of $\Bbb R$ onto $\Bbb R$, which is an open set. Thus, $f$ takes open sets to open sets, but $f$ is certainly not continuous.

The last part of (3) is also wrong: $\chi_{\left\{\frac1n:n\in\Bbb N\right\}}$ is bounded and has only countably many points of discontinuity, so it’s Riemann integrable.

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Thanks a lot!! I am a new user so I constantly mess things up. And the misunderstanding make me feel sad. Then my reputation drop so I can't even vote for a great answer. So I erased them to avoid getting a negative reputation...I really appreciate your help and effort! Thank you again! – user48601 Dec 16 '12 at 21:19
@Moriah: You’re welcome. One tip: in general people like it better if you type in the question, rather than link to an image (unless, of course, the question requires some sort of picture to make sense). I really don’t understand why so many people downvoted the question, though; if they’d actually looked at it, they’d have seen that it was a perfectly good one. – Brian M. Scott Dec 16 '12 at 21:22
I got what you said about the last two parts of (6). But I have one last question about last part of (3). I remember one theorem said that one function is Riemann integrable as long as its discontinuity forms a set of measure zero. We did't discuss Measure theory in detail in class. However, I used that theorem and I thought the interval [0,1]- the sequence 1/n is not countable. So it must not form a measure zero set. Then I got wrong. But where exactly did I get wrong? – user48601 Dec 16 '12 at 21:23
@Moriah: The set $\left\{\frac1n:n\in\Bbb N\right\}$ is countable, and the set of discontinuities is $\{0\}\cup\left\{\frac1n:n\in\Bbb N\right\}$, which is also countable, so both are sets of measure $0$. – Brian M. Scott Dec 16 '12 at 21:25
M.Scott: ! Thanks a lot. I am not quite clear about "the set of discontinuity" before – user48601 Dec 16 '12 at 21:29

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