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In each of the following, $f$ is assumed to be continuous. Pick out the cases when $f$ cannot be onto.

a. $f \colon [−1, 1] \to \mathbb{R}$.

b. $f \colon [−1, 1] \to\mathbb{Q}\cap [−1, 1]$.

c. $f \colon \mathbb{R} \to [−1, 1]$.

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closed as too localized by t.b., Michael Greinecker, William, Qiaochu Yuan Sep 15 '12 at 5:26

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5  
What have you tried? –  Matt Pressland Sep 11 '12 at 10:35
2  
Insights, ideas, background, effort done...? Have you checked anything about connectedness, compactness, boundness...? –  DonAntonio Sep 11 '12 at 10:38
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I picked out the cases where $f$ cannot be onto. Yay! –  Michael Greinecker Sep 11 '12 at 10:38
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thanks for your hints. now i can say that a is true as[-1,1] is compact and R is not. and b is also true as [-1,1] is connected and range is not. but what can i say about c? please help. –  poton Sep 11 '12 at 10:53
3  
Questions like part c. often give me a sinus headache. –  robjohn Sep 14 '12 at 1:25

1 Answer 1

up vote 0 down vote accepted

A function $ f: A \to B $ means, Domain of $ D(f) $ is A and range of the function $R(f) \subset B $

So, for a function to be surjective $ R(f) = B $

Give function is continuous.

  1. What is the image of bounded set under continuous map?
  2. What is the image of connected set under continuous map?
  3. What is the image of compact set under continuous map?

If you can give examples to above 3 you can answer the question.

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image of connected/compact set under continuous map is connected/compact. image of bounded set under continuous map may be bounded or may not be.i already got the answers for a & b but from this how can we conclude about c. –  poton Sep 11 '12 at 11:08
    
This approach will (usually) only allow you to disprove the existence of such a function. So having so far failed to do that, it might be worth spending a few minutes trying to think of an example of a continuous surjection $\mathbb{R}\to[-1,1]$. –  Matt Pressland Sep 11 '12 at 11:15
    
Is (0 1) isomorphic to R –  Ram Sep 11 '12 at 11:35

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