I am trying to prove: $A$ compact, $B$ closed $\Rightarrow A+B = \{a+b | a\in A, b\in B\}$ closed (exercise in Rudin's Functional Analysis), where $A$ and $B$ are subsets of a topological vector space $X$. In case $X=\mathbb{R}$ this is easy, using sequences. However, since I was told that using sequences in topology is "dangerous" (don't know why though), I am trying to prove this without using sequences (or nets, which I am not familiar with). Is this possible?

My attempt was to show that if $x\notin A+B$, then $x \notin \overline{A+B}$. In some way, assuming $x\in\overline{A+B}$ should then contradict $A$ being compact. I'm not sure how to fill in the details here though. Any suggestions on this, or am I thinking in the wrong direction here?

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    Using sequences is ok in metrizable spaces, in which the topological closure is the same as the sequential closure. In spaces which are not metrizable the sequential closure might be properly contained in the topological one. This is probably what they meant when they told you that using sequences "might be dangerous". – Giuseppe Negro Oct 5 '13 at 12:13
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    Here's an example of where using sequences can be "dangerous". Consider the TVS $X = \mathbb{R}^{[0,1]}$ (i.e. all functions $[0,1] \to \mathbb{R}$) with its product topology, and consider the linear subspace $E$ consisting of all those functions which are Borel measurable. Since a pointwise limit of measurable functions is measurable, the limit of any sequence from $E$ is again in $E$. From this you might think that $E$ is closed in $X$. But this is false: actually $E$ is dense in $X$. – Nate Eldredge Oct 5 '13 at 14:03
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    One way to see this is to remember that the product topology has a basis consisting of sets of the form $U = \prod_x U_x$, where $U_x$ is open in $\mathbb{R}$ and $U_x = \mathbb{R}$ for all but finitely many $x$. So let $x_1, \dots, x_n$ be those $x$ such that $U_x \ne \mathbb{R}$. Then we can say $f \in U$ iff $f(x_i) \in U_{x_i}$ for $i=1,\dots,n$. If for each $n$ we choose $y_i \in U_{x_i}$, we can certainly find a Borel function $f$ with $f(x_i) = y_i$ for each $i$. (For instance, we could let $f$ be a polynomial of degree $n$.) Thus $f \in U$, so $E$ is dense. – Nate Eldredge Oct 5 '13 at 14:08
  • @Nate Eldredge Thanks, that does clear things up. So iff there's a countable local base, using sequences is safe? What about those nets, are they safer? – ScroogeMcDuck Oct 5 '13 at 14:09
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    @ScroogeMcDuck: More or less. Roughly speaking, a space where you can do everything with sequences is called sequential. Every space with a countable local base is sequential, but not conversely (there's a counterexample on the Wikipedia page). Nets are safe: statements like "$E$ is closed iff every convergent net of points in $E$ has its limit in $E$" are true in every topological space. – Nate Eldredge Oct 5 '13 at 14:14
up vote 12 down vote accepted

Suppose $x\notin A+B$, then for each $a\in A$, $x \notin a+B$, which is a closed set (since $v \mapsto a+v$ is a homeomorphism). Since every TVS is regular, there are open sets $U_a$ and $V_a$ such that $$ x\in U_a, \quad a+B \subset V_a, \quad \text{ and } U_a\cap V_a = \emptyset $$ Now, $$ V_a - B = \cup_{b\in B} (V_a - b) $$ is open and contains $a$. Hence, $A\subset \cup_{a\in A}(V_a - B)$. Since $A$ is compact, there is a finite set $\{a_1, a_2, \ldots a_n\}$ such that $$ A \subset \cup_{i=1}^n (V_{a_i} - B) $$ Let $U = \cap_{i=1}^n U_{a_i}$, then $U$ is a neighbourhood of $x$.

We claim that $U\cap (A+B) = \emptyset$. If not, then $y = a+b \in U\cap(A+B)$, then $$ y \in V_{a_i} \quad\text{ for some } i $$ and $y \in U_{a_i}$, which is a contradiction.

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    Though $\mathbb{Q}$ is not even closed in $\mathbb{R}$, right? But it's clear, thank you very much. – ScroogeMcDuck Oct 5 '13 at 13:52
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    Would you please explain how you conclude, in the end, that $y=a+b$ should be in $V_{a_i}$ for some $i$ ? I agree that $a \in (V_{a_i} - B)$ for some $i$, but this does not (directly) imply that $a+b \in V_{a_i}$ for some $i$. – Nicolas Apr 9 '14 at 14:38
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    I agree with @Nicolas. I don't think the last contradiction holds. $y\in A+B$ implies only $y\in \left[\cup_{i=1}^n(V_{a_i}-B)\right]+B=\cup_{i=1}^n(V_{a_i}-B+B)$ which is not the same as $\cup_{i=1}^n V_{a_i}$. I think there are some issues with this last step of the proof. – Fang Jing Jun 8 '14 at 21:29
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    I know that this is an old post, but I answer the question of Nicolas and @FangJing so that other users can use it: We have $y \in \bigcup_{i=1}^n (V_{a_i} - B)$, so for some $i$, we have $y \in (V_{a_i} - B)$. Note that $V_{a_i} - B = \{ v \ : \ v \in V_{a_i}, \quad v \notin B \}$. Thus, $ V_{a_i} - B \subseteq V_{a_i}$, and then $y \in V_{a_i}$. – Hossein Moradi Jun 19 '15 at 3:50
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    @HosseinMoradi Isn't $V_{a_i}-B=\{v+(-b):v\in V_{a_i},\quad b\in B\}$? – Pedro Aug 24 '15 at 13:31

If $x\notin (A+B)$, then $A\cap(x-B)=\varnothing$. Since $(x-B)$ is closed, it follows from Theorem 1.10 in Rudin's book that there exists a neighborhood $V$ of $0$ such that $(A+V)\cap(x-B+V)=\varnothing$. Therefore $(A+B+V)\cap(x+V)=\varnothing$ and, in particular, $(A+B)\cap(x+V)=\varnothing.$ As $(x+V)$ is a neighborhood of $x$, this shows that $x\notin \overline{(A+B)}$. (Proof taken from Berge's book.)

  • Very good (+1). Maybe it would be more telling, for the hasty reader, if you mentioned where exactly compactness enters in the play. Something like [in Rudin's book that there exists a neighborhood]--->[in Rudin's book that, due to the compactness of $A$, there exists a neighborhood] – Duchamp Gérard H. E. Mar 12 '17 at 10:19

Do not hesitate to use nets and subnets because there is no loss of generality (i.e. they are sufficiently powerful to match the needs of general topology). In general, filters are suited when subsets or domains (as, for example, for germs of functions or asymptotic analysis) are under consideration, and nets are suited for computation of limits (which can be made the case here).

Proof Let $z$ be a boundary point of $A+B$, there exists a net $(z_\alpha)_{\alpha\in X}$ in $A+B$ which converges to $z$ (not yet known as belonging to $A+B$ and to be proved to be so) i.e. putting $z\in \lim_{\alpha \in X}z_\alpha$ in the non-Hausdorff case.

One writes $z_\alpha=x_\alpha+y_\alpha$ with $x_\alpha\in A,\ y_\alpha\in B$. As $A$ is compact, there exists a subnet $x_{\phi(\beta)}$ with acceptable $\phi:Y\rightarrow X$ which converges in $A$. Set $x=\lim_{\beta}x_{\phi(\beta)}$, from what precedes $y_{\phi(\beta)}=z_{\phi(\beta)}-x_{\phi(\beta)}$ converges to $z-x$ and this point lies in $B$ (as $B$ is closed). So $z=x+y\in A+B$. QED

  • Great you got it!!! :D Have a look also in our chat. ;) – C-Star-Puppy Jan 1 '17 at 21:14
  • Thanks ! I go to the chat right now :) – Duchamp Gérard H. E. Jan 1 '17 at 23:12
  • Thanks, now it even works in TVS with semi-norm (which aren't Hausdorff in general). – Cloudscape Mar 7 '17 at 8:49
  • @Cloudscape Are your compact Hausdorff ? (there are two definitions available on the market) – Duchamp Gérard H. E. Mar 8 '17 at 15:26
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    @Cloudscape You're right in intention: when I had in mind $\lim_{\alpha \in X}z_\alpha=z$, I'll make this precise in my answer by the explicit mention $z\in \lim_{\alpha \in X}z_\alpha$. Glad that you're a fan of Bourbaki, this is a very good (but inequal) treatise. I learned all the basic maths in Bourbaki when I was young. – Duchamp Gérard H. E. Mar 12 '17 at 9:45

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