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I know that any [a,b] are compact. However, if it is compact, then $f(\,[0,1]\,)$ with the continuous function $f(x)=\ln(x)$ would be compact too, but $f(\,[0,1]\,)=(-\infty,0]$ which is not compact. What is wrong with that?

For second question, I think $(0,1]=[0,1]$ and it is compact in $\mathbb R^{+}$ considering $f$ is defined on $\mathbb R^+$. But with the same logic in first question, it seems not compact. What is wrong with that?

Taha Direk
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    The log function isn’t defined at $0$: you’re actually looking at the image of $(0,1]$, which is not compact. – Brian M. Scott Dec 29 '20 at 18:24
  • $(0,1]=[0,1]$ is wrong. – Paul Frost Dec 29 '20 at 18:41
  • @BrianM.Scott thanks for your answer, I edited the question, can you help me with 2nd question too? – Taha Direk Dec 29 '20 at 18:41
  • @PaulFrost Aren't they equal in $\mathbb R^{+}$ – Taha Direk Dec 29 '20 at 18:42
  • No, $(0,1]$ is a subset of $\mathbb R^+$, but $[0,1]$ is not. – Paul Frost Dec 29 '20 at 18:43
  • @PaulFrost can't we say that $(0,1]$ includes all of its limit points in $\mathbb R^{+}$ ? (by closedness of $[0,1]$ in $\mathbb R$, it has all of limit points in $\mathbb R$) – Taha Direk Dec 29 '20 at 18:47
  • @TahaDirek: Yes, $(0,1]$ is a closed subset of $\Bbb R^+$, but that does not make it compact, and it certainly does not make it equal to $[0,1]$: $0$ is in $[0,1]$ and not in $(0,1]$. – Brian M. Scott Dec 29 '20 at 18:54
  • @BrianM.Scott I meant equal in $\mathbb R^{+}$ likewise $(-\sqrt2,\sqrt2)=[-\sqrt2,\sqrt2]$ in $\mathbb Q$, isn't it correct? I was thinking that closedness and boundedness imply compactness, but realized that it isn't, thanks for that – Taha Direk Dec 29 '20 at 19:00
  • @TahaDirek: That is an improper use of the word equal. What you mean is that they have the same intersection with $\Bbb R^+$. No, it is not correct to say that $(-\sqrt2,\sqrt2)=[-\sqrt2,\sqrt2]$ in $\Bbb Q$; what is correct is to say that $(-\sqrt2,\sqrt2)\cap\Bbb Q=[-\sqrt2,\sqrt2]\cap\Bbb Q$. – Brian M. Scott Dec 29 '20 at 19:03
  • @BrianM.Scott: So, we can't define $(0,1]$ as a closed interval, thus my conclusion of $(0,1]$ is compact in $\mathbb R^{+}$ is wrong, is it correct? – Taha Direk Dec 29 '20 at 19:20
  • @TahaDirek: That is correct: a closed interval includes both of its endpoints. – Brian M. Scott Dec 29 '20 at 19:24
  • @BrianM.Scott thanks for your effort, all the best :) – Taha Direk Dec 29 '20 at 19:26
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    @TahaDirek: You’re welcome! – Brian M. Scott Dec 29 '20 at 19:32
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    Compactness is an absolute property of a space $X$, it does not depend on a bigger ambient space $Y \supset X$. The sets $U_n = (1/n,1]$ form an (infinite) open cover of $(0,1]$, but there does not exist a finite subcover. Hence $(0,1]$ is not compact. – Paul Frost Dec 29 '20 at 23:06
  • @PaulFrost I got the idea, thanks for all :) – Taha Direk Dec 30 '20 at 10:23

2 Answers2

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The problem is that $\ln$ is only defined on $(0,\infty)$, thus it does not make sense to consider $f([0,1])$.

Moreover, we have $\lim_{x \to 0} \ln(x) = -\infty$, therefore there is no continuous extension of $\ln$ to a map $[0,\infty) \to \mathbb R$. But we can work with the extended real line $\bar {\mathbb R} = \mathbb R \cup \{-\infty, +\infty \}$ and topologize it in the well-known way. Note that $\bar {\mathbb R}$ is compact. Then $\ln$ extends continuously to $\ln : [0,\infty) \to \bar {\mathbb R}$ by defining $\ln(0) = -\infty$. Now $\ln([0,1]) = [-\infty,0]$ which is compact.

Paul Frost
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You cannot take $\log([0,1]$, since it is not well defined in zero.

But if you take $\log([\varepsilon,1]$, with $\epsilon>0$, then you get a compact interval as image since logarithm is continuous in its domain.

Son Gohan
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