choose the coorect option $1)$ $K$ is uncountable $2)$ $K$ is finite.












0












$begingroup$


Let $Ksubset mathbb{C}$ be a bounded set. Let $H(mathbb{C})$ denote the set of all entire functions and let $C(K)$ denote the set of all continuous functions on $K$. Consider the restriction map $r:H(mathbb{C})rightarrow C(K)$ by $r(f)=f|_K.$ Then $r$ is injective if



then choose the coorect option



$1)$ $K$ is uncountable



$2)$ $K$ is finite.



My attempt : $K$ must be finite because $K$ is bounded as we know that bounded set are generally countable set so , option $1)$ is false



Is its True ?



Any hints/solution will be appreciated



thanks u






complex-analysis







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Why are bounded sets countable?
    $endgroup$
    – anomaly
    Dec 1 '18 at 22:39










  • $begingroup$
    bounded set is finite so it can be countable @anomaly
    $endgroup$
    – jasmine
    Dec 1 '18 at 22:40






  • 1




    $begingroup$
    No, a bounded set is not necessarily finite.
    $endgroup$
    – anomaly
    Dec 1 '18 at 23:32
















0












$begingroup$


Let $Ksubset mathbb{C}$ be a bounded set. Let $H(mathbb{C})$ denote the set of all entire functions and let $C(K)$ denote the set of all continuous functions on $K$. Consider the restriction map $r:H(mathbb{C})rightarrow C(K)$ by $r(f)=f|_K.$ Then $r$ is injective if



then choose the coorect option



$1)$ $K$ is uncountable



$2)$ $K$ is finite.



My attempt : $K$ must be finite because $K$ is bounded as we know that bounded set are generally countable set so , option $1)$ is false



Is its True ?



Any hints/solution will be appreciated



thanks u






complex-analysis







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Why are bounded sets countable?
    $endgroup$
    – anomaly
    Dec 1 '18 at 22:39










  • $begingroup$
    bounded set is finite so it can be countable @anomaly
    $endgroup$
    – jasmine
    Dec 1 '18 at 22:40






  • 1




    $begingroup$
    No, a bounded set is not necessarily finite.
    $endgroup$
    – anomaly
    Dec 1 '18 at 23:32














0












0








0





$begingroup$


Let $Ksubset mathbb{C}$ be a bounded set. Let $H(mathbb{C})$ denote the set of all entire functions and let $C(K)$ denote the set of all continuous functions on $K$. Consider the restriction map $r:H(mathbb{C})rightarrow C(K)$ by $r(f)=f|_K.$ Then $r$ is injective if



then choose the coorect option



$1)$ $K$ is uncountable



$2)$ $K$ is finite.



My attempt : $K$ must be finite because $K$ is bounded as we know that bounded set are generally countable set so , option $1)$ is false



Is its True ?



Any hints/solution will be appreciated



thanks u






complex-analysis







share|cite|improve this question









$endgroup$




Let $Ksubset mathbb{C}$ be a bounded set. Let $H(mathbb{C})$ denote the set of all entire functions and let $C(K)$ denote the set of all continuous functions on $K$. Consider the restriction map $r:H(mathbb{C})rightarrow C(K)$ by $r(f)=f|_K.$ Then $r$ is injective if



then choose the coorect option



$1)$ $K$ is uncountable



$2)$ $K$ is finite.



My attempt : $K$ must be finite because $K$ is bounded as we know that bounded set are generally countable set so , option $1)$ is false



Is its True ?



Any hints/solution will be appreciated



thanks u






complex-analysis




complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 22:37









jasminejasmine

1,689416




1,689416








  • 2




    $begingroup$
    Why are bounded sets countable?
    $endgroup$
    – anomaly
    Dec 1 '18 at 22:39










  • $begingroup$
    bounded set is finite so it can be countable @anomaly
    $endgroup$
    – jasmine
    Dec 1 '18 at 22:40






  • 1




    $begingroup$
    No, a bounded set is not necessarily finite.
    $endgroup$
    – anomaly
    Dec 1 '18 at 23:32














  • 2




    $begingroup$
    Why are bounded sets countable?
    $endgroup$
    – anomaly
    Dec 1 '18 at 22:39










  • $begingroup$
    bounded set is finite so it can be countable @anomaly
    $endgroup$
    – jasmine
    Dec 1 '18 at 22:40






  • 1




    $begingroup$
    No, a bounded set is not necessarily finite.
    $endgroup$
    – anomaly
    Dec 1 '18 at 23:32








2




2




$begingroup$
Why are bounded sets countable?
$endgroup$
– anomaly
Dec 1 '18 at 22:39




$begingroup$
Why are bounded sets countable?
$endgroup$
– anomaly
Dec 1 '18 at 22:39












$begingroup$
bounded set is finite so it can be countable @anomaly
$endgroup$
– jasmine
Dec 1 '18 at 22:40




$begingroup$
bounded set is finite so it can be countable @anomaly
$endgroup$
– jasmine
Dec 1 '18 at 22:40




1




1




$begingroup$
No, a bounded set is not necessarily finite.
$endgroup$
– anomaly
Dec 1 '18 at 23:32




$begingroup$
No, a bounded set is not necessarily finite.
$endgroup$
– anomaly
Dec 1 '18 at 23:32










1 Answer
1






active

oldest

votes


















2












$begingroup$

HINT



First, why your attempt is wrong. $K$ could be infinite and uncountable, take the set of real numbers $[0, 1]$ for example.



Because it's asking if $r$ is injective, you could take a look at the cardinality of the sets $H(mathbb{C})$ and $C(K)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    can u give me the example of $H( mathbb{C})$ ?
    $endgroup$
    – jasmine
    Dec 1 '18 at 23:22






  • 1




    $begingroup$
    I don't actually know how you would calculate the cardinality of $H(mathbb{C})$, but we can safely say it's infinite. (1) If $K$ is uncountable, $C(K)$ would be infinite. (2) If $K$ is finite, $C(K)$ would be finite as well since it's just the set of all permutations of $K$. Obviously if $C(K)$ is finite $r$ is definitely not an injection, so choose (1)
    $endgroup$
    – EvanHehehe
    Dec 2 '18 at 0:12











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

HINT



First, why your attempt is wrong. $K$ could be infinite and uncountable, take the set of real numbers $[0, 1]$ for example.



Because it's asking if $r$ is injective, you could take a look at the cardinality of the sets $H(mathbb{C})$ and $C(K)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    can u give me the example of $H( mathbb{C})$ ?
    $endgroup$
    – jasmine
    Dec 1 '18 at 23:22






  • 1




    $begingroup$
    I don't actually know how you would calculate the cardinality of $H(mathbb{C})$, but we can safely say it's infinite. (1) If $K$ is uncountable, $C(K)$ would be infinite. (2) If $K$ is finite, $C(K)$ would be finite as well since it's just the set of all permutations of $K$. Obviously if $C(K)$ is finite $r$ is definitely not an injection, so choose (1)
    $endgroup$
    – EvanHehehe
    Dec 2 '18 at 0:12
















2












$begingroup$

HINT



First, why your attempt is wrong. $K$ could be infinite and uncountable, take the set of real numbers $[0, 1]$ for example.



Because it's asking if $r$ is injective, you could take a look at the cardinality of the sets $H(mathbb{C})$ and $C(K)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    can u give me the example of $H( mathbb{C})$ ?
    $endgroup$
    – jasmine
    Dec 1 '18 at 23:22






  • 1




    $begingroup$
    I don't actually know how you would calculate the cardinality of $H(mathbb{C})$, but we can safely say it's infinite. (1) If $K$ is uncountable, $C(K)$ would be infinite. (2) If $K$ is finite, $C(K)$ would be finite as well since it's just the set of all permutations of $K$. Obviously if $C(K)$ is finite $r$ is definitely not an injection, so choose (1)
    $endgroup$
    – EvanHehehe
    Dec 2 '18 at 0:12














2












2








2





$begingroup$

HINT



First, why your attempt is wrong. $K$ could be infinite and uncountable, take the set of real numbers $[0, 1]$ for example.



Because it's asking if $r$ is injective, you could take a look at the cardinality of the sets $H(mathbb{C})$ and $C(K)$.






share|cite|improve this answer











$endgroup$



HINT



First, why your attempt is wrong. $K$ could be infinite and uncountable, take the set of real numbers $[0, 1]$ for example.



Because it's asking if $r$ is injective, you could take a look at the cardinality of the sets $H(mathbb{C})$ and $C(K)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 '18 at 23:32









anomaly

17.5k42664




17.5k42664










answered Dec 1 '18 at 23:13









EvanHeheheEvanHehehe

1247




1247












  • $begingroup$
    can u give me the example of $H( mathbb{C})$ ?
    $endgroup$
    – jasmine
    Dec 1 '18 at 23:22






  • 1




    $begingroup$
    I don't actually know how you would calculate the cardinality of $H(mathbb{C})$, but we can safely say it's infinite. (1) If $K$ is uncountable, $C(K)$ would be infinite. (2) If $K$ is finite, $C(K)$ would be finite as well since it's just the set of all permutations of $K$. Obviously if $C(K)$ is finite $r$ is definitely not an injection, so choose (1)
    $endgroup$
    – EvanHehehe
    Dec 2 '18 at 0:12


















  • $begingroup$
    can u give me the example of $H( mathbb{C})$ ?
    $endgroup$
    – jasmine
    Dec 1 '18 at 23:22






  • 1




    $begingroup$
    I don't actually know how you would calculate the cardinality of $H(mathbb{C})$, but we can safely say it's infinite. (1) If $K$ is uncountable, $C(K)$ would be infinite. (2) If $K$ is finite, $C(K)$ would be finite as well since it's just the set of all permutations of $K$. Obviously if $C(K)$ is finite $r$ is definitely not an injection, so choose (1)
    $endgroup$
    – EvanHehehe
    Dec 2 '18 at 0:12
















$begingroup$
can u give me the example of $H( mathbb{C})$ ?
$endgroup$
– jasmine
Dec 1 '18 at 23:22




$begingroup$
can u give me the example of $H( mathbb{C})$ ?
$endgroup$
– jasmine
Dec 1 '18 at 23:22




1




1




$begingroup$
I don't actually know how you would calculate the cardinality of $H(mathbb{C})$, but we can safely say it's infinite. (1) If $K$ is uncountable, $C(K)$ would be infinite. (2) If $K$ is finite, $C(K)$ would be finite as well since it's just the set of all permutations of $K$. Obviously if $C(K)$ is finite $r$ is definitely not an injection, so choose (1)
$endgroup$
– EvanHehehe
Dec 2 '18 at 0:12




$begingroup$
I don't actually know how you would calculate the cardinality of $H(mathbb{C})$, but we can safely say it's infinite. (1) If $K$ is uncountable, $C(K)$ would be infinite. (2) If $K$ is finite, $C(K)$ would be finite as well since it's just the set of all permutations of $K$. Obviously if $C(K)$ is finite $r$ is definitely not an injection, so choose (1)
$endgroup$
– EvanHehehe
Dec 2 '18 at 0:12


















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