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If $A$ is a subset of $\Bbb R$ and $f : A \to \Bbb R$ is a continuous function then prove or disprove the following.

  1. If $A$ is a bounded set but not closed then $f(A)$ is bounded.

  2. If $A$ is a closed set but not bounded then $f(A)$ is closed.

My Attempt : I have seen a result which tells that Image of closed and bounded set is closed and bounded. Will happen when one condition either closed or bounded is relaxed.

Norse
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Baljeet
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2 Answers2

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A brief answer to settle one of the questions that Baljeet asks: in general, a continuous function $f: \mathbb{R} \to \mathbb{R}$ will not be closed (i.e. enjoying the property of sending closed subsets to closed subsets). Here is one possible counter-example: consider any $a \in (0,1)$ and take $f$ to be the exponential function of base $a$, in other words such that

$$f: \mathbb{R} \to \mathbb{R} \\ f(x)=a^x$$

Clearly $\mathbb{N} \subset \mathbb{R}$ is a closed subset in the order topology (the standard one at work in analysis), however $f(\mathbb{N})=\{a^n\}_{n \in \mathbb{N}}$ is no longer closed: since $$a^n \xrightarrow{n \to \infty} 0$$

we infer that $0$ is a point adherent to this subset (i.e. in its closure) lying though in the complement of this subset.

A perhaps even clearer counter-example can be found in the instance of $$g: \mathbb{R} \to \mathbb{R} \\ g(x)=\frac{1}{x^2+1}$$

Indeed $g([0, \infty))=(0,1]$ is not closed.

ΑΘΩ
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1) Take $A=(0,\pi/2)$ Then $f(x)=\tan(x)$ is unbounded, but $A$ is bounded.

2) Take $A=\Bbb R$ then $A$ is closed (though also open) and the map $f(x)=\frac{1}{1+e^{-x}}$ has $f(A)=(0,1)$ which is open.

Norse
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  • the general idea of your counterexamples is in the right spirit, but there are bothersome errors in both your answers 1) and 2). – ΑΘΩ Jan 06 '20 at 12:58
  • @ΑΘΩ oh yeah! Thank you, those are obvious errors – Norse Jan 06 '20 at 13:16
  • @Norse, Is f(A) bounded for all bounded subsets A of R ? – Baljeet Jan 06 '20 at 13:28
  • @Baljeet not too sure what you mean? I just gave an example where this is not true – Norse Jan 06 '20 at 13:29
  • I saw somewhere, If f : R -> R is continuous then f(A) is bounded for all bounded subsets A of R. How is it possible ? – Baljeet Jan 06 '20 at 13:39
  • @Baljeet Yeah, that is true – Norse Jan 06 '20 at 14:01
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    @Baljeet: that will in general be true since 1) bounded subsets of $\mathbb{R}$ are relatively compact (this means in general that they are included in a certain compact subset) 2) compactness is preserved under continuous maps and 3) compact subsets of $\mathbb{R}$ (or more generally a metric space) will necessarily be bounded. – ΑΘΩ Jan 06 '20 at 14:10
  • @Norse, If f : R -> R is continuous function. Then f(A) is closed for all closed subsets A of R. Is it true in general ? – Baljeet Jan 06 '20 at 14:29
  • @Norse,Can you provide me a solution of above results in general ? – Baljeet Jan 06 '20 at 14:37
  • @Baljeet I think not. But I am not entirely sure – Norse Jan 06 '20 at 14:43
  • @ΑΘΩ, Can you provide me a solution or a link of the following ? – If f : R -> R is continuous function. Then f(A) is closed for all closed subsets A of R. Is it true in general ? – Baljeet Jan 06 '20 at 14:54
  • @Baljeet: to make things clear and explicit, let me present a full answer to this question that you pose; please refer to it below. – ΑΘΩ Jan 07 '20 at 00:58