Show that $c_0 \subset \ell^\infty$ is a closed subspace.

I'm trying to validate my proof which went as follows.

Let $(x_n^k)_{k=1}^\infty$ be a sequence in $c_0$ such that $\lim_{k \to \infty} x_n^k = a$. This implies that for all $\varepsilon > 0$ there exists $k_0 \in \Bbb N$ such that $k > k_0$ implies $$|x_n^k-a| \le \|x_n^k-a\|_\infty < \varepsilon$$

thus $$-\varepsilon < x_n^k -a < \varepsilon \iff \varepsilon -x_n^k > a > -\varepsilon +x_n^k.$$

However since $x_n^k \in c_0$ for all $k$ it means that for a fixed $k$ we have that $\lim_{n \to \infty} x_n^k = 0$.

Thus taking limits on $$\varepsilon -x_n^k > a > -\varepsilon +x_n^k$$ with respect to $n$ we get that $$-\varepsilon < a <\varepsilon$$ for all $k$.

But this is the same as saying that $$|a|<\varepsilon$$ which would mean that $a$ converges to $0$ and is thus in $c_0$.

The part with taking limits both sides is where I feel like I am not being very rigorous is there a better way to conclude that?

  • $\begingroup$ If $\mathbf{x}_n$ is a sequence in $c_0$ that converges to some $\mathbf{x}\in\ell_\infty$ then for $\varepsilon >0$, there is $N$ such that $\|\mathbf{x}_n-\mathbf{x}\|_\infty<\varepsilon/2$. As $\mathbf{x}_N\in c_0$, there is $M$ such that $m\geq M$ implies that $|\mathbf{x}_N(m)|<\varepsilon/2$ for all $m\geq M$. Hence $|\mathbf{x}(m)|\leq \|\mathbf{x}-\mathbf{x}_N\|_\infty+|\mathbf{x}_N(m)|<\varepsilon$ for all $m\geq M$. $\endgroup$ Oct 22, 2022 at 22:59
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    $\begingroup$ It's a little strange to write $\lim_{k \to \infty} x^k_n = a$. Shouldn't the right side depend on $n$? A similar issue occurs with the line $|x^k_n - a| \leq \Vert x^k_n - a \Vert_{\infty}$, which also looks strange. It looks like your plan is to start with a sequence in $c_0$ that converges to a limit, and showing that that limit lives in $c_0$. Here, "converges to a limit" means "converges in the $\ell^\infty$ norm." $\endgroup$ Oct 22, 2022 at 23:08