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Let $(X,\|\;\|)$ be a normed vector space over $K$ and let $A,B\subseteq X$ be s.t. $A$ is compact and $B$ is closed. I want to prove that:

$$(\alpha A+\beta B)\;\text{ is closed }\;\forall \alpha,\beta\in K$$ So, given our function $\odot$ as in here, my attempt goes like this:

Let $\alpha,\beta\in K$.

If $\alpha=\beta=0$, clearly $$\alpha A+\beta B=\{0\}\text{ is closed}$$ Now, since $\odot$ is continuous and $A$ is compact, we have that $\;\forall \alpha\neq0,\;\alpha A=\odot(\alpha,A)\;$ is compact and thus closed.

Then, let $\;y\in \beta B$ so, $$ y=\beta b_y $$ for some $b_y\in B$. But, since $B$ is closed, there exists $\{b_n\}_{n\in\Bbb N}\subset B$ s.t. $b_n\xrightarrow[n\to\infty]{}b_y$
So let $y_n=\beta b_n\;\;\forall n\in\Bbb N$. Thus, $\{y_n\}_{n\in\Bbb N}\subset \beta B$ and $\|y_n-y\|=\|\beta b_n-\beta b_y\|=|\beta|\|b_n-b_y\|\xrightarrow[n\to\infty]{}0$
Thus, we get that $$\forall y\in\beta B\;\exists\;\{y_n\}_{n\in\Bbb N}\subset \beta B \text{ s.t. } y_n\xrightarrow[n\to\infty]{}y$$ so $\beta B$ is closed $\;\forall \beta\in K$.

Now, $$ \text{Let }x\in (\alpha A+\beta B)\Rightarrow\;\exists\;a^*\in\alpha A\;\land\;b^*\in\beta B\text{ s.t. } x=a^*+b^* $$ so, since $\alpha A$ and $\beta B$ are closed, $$ \exists\;\{a_n^*\}_{n\in\Bbb N}\subset\alpha A\text{ s.t. }a_n^*\xrightarrow[n\to\infty]{}a^*\\ \exists\;\{b_n^*\}_{n\in\Bbb N}\subset\beta B\text{ s.t. }b_n^*\xrightarrow[n\to\infty]{}b^* $$ so, taking $x_n=a_n^*+b_n^*\;\forall n\in\Bbb N$ is clear that $x_n\in (\alpha A+\beta B)\;\forall n\in\Bbb N$ and $$ \|x_n-x\|=\|a_n^*+b_n^*-a^*-b^*\|\le\|a_n^*-a^*\|+\|b_n^*-b^*\|\xrightarrow[n\to\infty]{}0 $$ so $x_n\xrightarrow[n\to\infty]{}x$ and, since this happens $\;\forall x\in (\alpha A+\beta B)$, thereby $(\alpha A+\beta B)$ is closed $\;\forall \alpha ,\beta\in K$

What do you think? Did I miss something?

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Arnulf
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1 Answers1

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You do not start correctly when you want to prove that a set is closed. When you want to prove that $A\subset X$ is closed, don't choose $x\in A$ and show that there is a sequence $(x_n)$ in $A$ that converges to $x$, since this sequence might be $x_n = x$. In contrast, you should start with a convergent sequence $(x_n)$ in $A$ and show that its limit also is an element of $A$.

In your case, $\beta B$ is closed. Clearly, this holds for $\beta = 0$. So, let $\beta\neq 0$. Choose $(x_n)\subset\beta B$, $x_n\to x$. Then $x_n = \beta b_n$ with $b_n\in B$. Now, $b_n = \beta^{-1}x_n\to\beta^{-1}x$. As $B$ is closed, $\beta^{-1}x\in B$, so $x\in\beta B$. Done.

Here, as an alternative, you can also define $f : X\to X$ by $f(x) := \beta^{-1}x$. Then $\beta B = f^{-1}(B)$. And that was it since $f$ is continuous.

Your proof that $\alpha A$ is compact is fine. Now, apply the above approach to show that $C + D$ is closed if $C$ is compact and $D$ closed. Use here that each sequence in $C$ has a convergent subsequence.

Friedrich Philipp
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  • I didn't use the compactness and I think something might also be wrong with that, but can't see how to use that $\alpha A$ is compact – Arnulf Mar 07 '16 at 00:59
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    You have a compact set $C = \alpha A$ and a closed set $D = \beta B$. You want to show that $C+D$ is closed. Thus, choose $(x_n)\subset C+D$ such that $x_n\to x$. Now, $x_n = c_n + d_n$, where $(c_n)\subset C$ and $(d_n)\subset D$. Now, use that $(c_n)$ has a convergent subsequence $(c_{n_k})$. What happens to $d_{n_k}$? – Friedrich Philipp Mar 07 '16 at 01:03
  • I can't see how $(c_{n_k})$ being convergent could help to show that $x\in C+D$ – Arnulf Mar 07 '16 at 01:37
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    You are too impatient. $d_{n_k} = x_{n_k} - c_{n_k}$. What happens when $k\to\infty$? – Friedrich Philipp Mar 07 '16 at 01:55
  • So, given that $c_{n_k}\xrightarrow[k\to\infty]{}c_0;(\in C)$ we get ${d_{n_k}}{k\in\Bbb N},$ a subsequence of ${d_n}{n\in\Bbb N}$, s.t. $;\forall k\in\Bbb N;;d_{n_k}=x_{n_k}-c_{n_k}$, for any ${x_{n_k}}{k\in\Bbb N}$ sub sequence of ${x_n}{n\in\Bbb N}$, and $b_{n_k}\xrightarrow[k\to\infty]{}x-c_0$. But $D$ is closed, so $x-c_0\in D$ which implies that $x\in c_0+D$ – Arnulf Mar 07 '16 at 02:11
  • Exactly. I guess you meant to write $d_{n_k}\to c - c_0$ (not $b_{n_k}$). In the end, $x\in c_0 + D$, so $x\in C+D$. Done. :o) – Friedrich Philipp Mar 07 '16 at 02:22
  • Yeap hehe. Thank you! – Arnulf Mar 07 '16 at 02:24