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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?

Martin Sleziak
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ScroogeMcDuck
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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 05 '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 05 '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 05 '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 05 '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 05 '13 at 14:14
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    Nets can sometimes be tricky though: for example, every sequence is a net, but a subnet of such a net need not be a sequence. – Nate Eldredge Oct 05 '13 at 14:15

4 Answers4

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If $x\notin (A+B)$, then $A\cap(x-B)=\varnothing$. Since $(x-B)$ is closed and $A$ is compact, 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.)

Kadmos
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Pedro
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    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
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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.

Prahlad Vaidyanathan
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  • Thanks, very clear. I am not completely sure where you used compactness though. Clearly you used it to when stating $A$ has a finite subcover ${V_{a_i}-B|i\in {1,2,...,n}}$, but if you would just use the cover ${V_{a}-B|a\in A}$, where would the argument fail? Is it maybe true that $U$ would then not necessarily be an open neighborhood? – ScroogeMcDuck Oct 05 '13 at 13:42
  • Yes, you need that to take a finite intersection of the $U_{a_i}$'s. For instance, if $A=\mathbb{Q}, B = \mathbb{Z}$, then $A+B$ is no longer closed in $\mathbb{R}$! – Prahlad Vaidyanathan Oct 05 '13 at 13:44
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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 05 '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 09 '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. – tvk Jun 08 '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 03:50
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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
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    That's why people should stop using the notation $A-B$ for ${x: x\in A, x \notin B}$. – Integral Aug 26 '15 at 17:12
  • @PrahladVaidyanathan If $X$ is a normed vector space, can we conclude that $A+B$ is compact? – John Mars Apr 28 '21 at 13:55
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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 $x=\lim_{\beta}x_{\phi(\beta)}$ ($x\in\lim_{\beta}x_{\phi(\beta)}$ in the non-Hausdorff case). 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

Duchamp Gérard H. E.
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  • @Cloudscape Are your compact Hausdorff ? (there are two definitions available on the market) – Duchamp Gérard H. E. Mar 08 '17 at 15:26
  • No, I usually use the non-Bourbaki ("open-cover") definition. But I proved every theorem on nets used above from scratch, and it worked out for me! Maybe I made a mistake though, although I hope not? – Cloudscape Mar 09 '17 at 11:46
  • I did use plenty of choice and contradiction though. – Cloudscape Mar 09 '17 at 11:48
  • @DuchampGérardH.E. – Cloudscape Mar 09 '17 at 11:57
  • @Cloudscape Yes, it is what Bourbaki calls quasi-compact. Note that in the MO question, there is no mention or use of a norm, neither semi-norms and, in my answer, of Hausdorff property so that the statement holds for TVS even non locally convex. – Duchamp Gérard H. E. Mar 12 '17 at 03:19
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    @DuchampGérardH.E. Yes; I would've assumed that there is some implicit Hausdorff assumption in place, also because of the theorem mentioned by Pedro in the other answer, and they are using the same book. Let me mention though that I'm a big fan of Bourbaki, and I also use filters and in particular (with exception of the empty uniform space, where we wouldn't have a filter) uniform structures, as given in Bourbaki. (On the other hand, for TVS I find Grothendieck's exposition superior, with the exception of some details which are only found in the Bourbaki book.) – Cloudscape Mar 12 '17 at 08:16
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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 09:45
  • @C-star-W-star [Great you got it!!! :D Have a look also in our chat. ;)] ---> I went to the chat years ago. Is it still accessible? – Duchamp Gérard H. E. Jan 25 '22 at 02:06
  • Remember that we were chatting about it but not sure it is still available, do you need something specific? – C-star-W-star Jan 25 '22 at 15:50
  • Not specific, I just wanted to browse into it. Is there a link to go there ? – Duchamp Gérard H. E. Jan 26 '22 at 14:47
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Here is another proof inspired by the sequential proof in the $\mathbb{R}^n$ case.

Let $x\in \overline{A + B}$. Let us show that it is in fact in $A+B$.

I assume the topological vector space $X$ is Hausdorff so that $\displaystyle \bigcap_{i\in I} V_i = \{x\}$ where $V_i$ runs over all neighborhoods of $x$.

By definition of being in the adherence, $\forall\ i\in I,\enspace V_i \cap (A+B) \neq \emptyset$. Let us introduce the compact subset $$K_i := \overline{\left\lbrace a \in A,\ \exists\, b\in B,\ a+b \in V_i \right\rbrace } \subset A$$ (compact as a closed subset of the compact subset $A$).

If $\ \displaystyle \bigcap_{i\in I} K_i = \emptyset $ then by the finite intersection property of compact sets, $\exists\ J \subset I,\ J$ finite such that $\ \displaystyle\bigcap_{j\in J} K_j = \emptyset $ but this means that $$ \left(\bigcap_{j\in J} V_j \right)\cap (A+B)= \emptyset $$ but since a finite intersection of neighborhoods of $x$ is also a neighborhood of it, this contradicts the fact that $x\in \overline{A + B}$. Hence $\ \displaystyle \bigcap_{i\in I} K_i \neq \emptyset ,\ \exists\ a_0\in A$ s.t. $\forall\ i\in I,\ \exists\ b_i \in B,\ a_0 + b_i \in V_i$.

One may say that the function $i \mapsto a_0 + b_i $ has limit $x$ "along the filter $I$" and thus that $$\lim_{\overset{\longrightarrow}{I}} b_i = x - a_0=: b_0 \in B$$ since the latter is closed. Finally, one does have $x=a_0 + b_0 $ with $a_0 \in A,\ b_0\in B$.

Comment:

  • I was interested in a (general if possible) proof of this statement while reading the proof of Thm 1.7 p.7 of "Functional Analysis, Sobolev Spaces and Partial Differential Equations" (UTX, 2011) by H. Brezis, or while reading (ii) of the proof of Proposition 1.3 p.69 of "Eléments de distributions et d'équations aux dérivées partielles" by C. Zuily
  • In fact there is a sequential proof for a slightly more general case in "Eléments de distributions et d'équations aux dérivées partielles" by C. Zuily, Proposition 6.3 p.79
Noix07
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