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Let $\tau:c_0\rightarrow\ell_\infty$ be the inclusion map. Let $x_0=\{1,1,\cdots\}$. Then $x_0\in\ell_\infty-c_0$. Since $\tau(x)=x\in c_0$ for all $x\in c_0$, there is no $x\in c_0$ such that $\tau(x)=x_0$. Hence, $\tau:c_0\rightarrow\ell_\infty$ is not surjective.

But my professor told me my logic is wrong on this proof. Can any one help me on this? Thank you!

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  • $\begingroup$ This isn't a proof (or, at least, it's unnecessarily verbose for one); you're just asserting that $x_0\in \ell^\infty$ has $x_0\notin c_0$, which is exactly what you're trying to prove. Still, there's not much to say beyond noting that (as a sequence) $x_0$ is bounded but doesn't converge to $0$. If I were doing this problem, I'd just simply write, "The sequence $x_0\in (1, 1, \dots)$ clearly lies in $\ell^\infty$ but not $c_0$"; there's nothing more useful to say. $\endgroup$ – anomaly Jun 26 '18 at 21:48
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Your proof is perfectly correct. Perhaps your professor expects you to show more details in some steps, such as when you claim that $x_0\in\ell_\infty-c_0$. You could also spell out more explicitly why there is no $x\in c_0$ such that $\tau(x)=x_0$ (namely, that $\tau(x)=x$ so this would mean $x_0=x\in c_0$ but $x_0\not\in c_0$).

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  • $\begingroup$ Thanks for your answer! Do you find anything unclear on the logic part? $\endgroup$ – Answer Lee Jun 27 '18 at 2:21
  • $\begingroup$ Not sure what you mean by "the logic part", but it seems fine to me. $\endgroup$ – Eric Wofsey Jun 27 '18 at 2:27
  • $\begingroup$ Because my professor said my proof is logically wrong and I didn't understand the definition of non-surjectivity at all. $\endgroup$ – Answer Lee Jun 27 '18 at 2:36
  • $\begingroup$ It sounds like your professor misunderstood your proof. $\endgroup$ – Eric Wofsey Jun 27 '18 at 2:37

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