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I know that the following exercise you can find on internet maybe with solution too, but I want to know, if my "solutions" are correct.

Let $X,Y\subset \mathbb{R}^n$, $X+Y=\{x+y;x\in X, y\in Y\}$. Prove it or find a counterexample.

1) If X, Y open, then X+Y is open

2) If X, Y closed, then X+Y is closed

3) If X, Y compact, then X+Y is compact

My ideas:
1) is true. Could you have a look if my solution is correct? My try: We know that for all $x\in X \exists \epsilon >0: B_{\epsilon}(x)=\{b\in X; \|b-x\|<\epsilon\}\subset X$, and for all $y\in Y \exists \epsilon' >0: B_{\epsilon'}(x)=\{b'\in X; \|b'-x\|<\epsilon'\}\subset Y$, because $X$ and $Y$ are open. Now: $\|z-(x+y)\|=\|z-x-y\|=\|z'-z'+z-x-y\|=\|z'-x+z-z'-y\|\le\|z'-x\|+\|z-z'-y\|<\epsilon+\epsilon'$. Define $\eta =\epsilon +\epsilon'$. We have: for $x+y\in X+Y \exists \eta >0, \eta =\epsilon +\epsilon': B_{\eta}(x+y)=\{z\in X+Y; \|z-(x+y)\|<\eta\}\subset X+Y$.

I'm not sure, if this is correct. If my "solution" is false, could you help to correct?

2) First I said that this is true, but after googeling I found out that you can find a counterexample. Could you help me to find the mistake of my "solution"? My try: For every sequence $(x_n)\subseteq X$ such that $x_n\to x_0$, $x_0\in \mathbb{R}^n$, it is $x_0\in X$, because X is closed. For every sequence $(y_n)\subseteq Y$, such that $y_n\to y_0$, $y_0\in \mathbb{R}^n$, it is $y_0\in Y$, because Y is closed.

Let $(z_n)\subseteq X+Y$ be a sequence, $z_n=x_n+y_n$ for every $n\in\mathbb{N}$ and let $z_n\to z_0 \in \mathbb{R}^n$. But it is $z_n=x_n+y_n\to x_0+y_0$ and by the uniqueness of limits $z_0=x_0+y_0\in X+Y$, because X and Y are closed.

But my try has to be wrong I think, because you find counterexamples for 2). And I don't find the mistake :(, could you help me?

3) Is it correct? I would say yes. Maybe I can prove it with a continuous function and $X+Y$ as it's compact image.

Regards

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    $\begingroup$ 2) The worry is that there's some sequence $(x_n+y_n)$ that converges, but not to a point in $X+Y$. As you've shown, if $(x_n)$ and $(y_n)$ converge, then this sequence must converge to a point in $X+Y$. But what if neither $(x_n)$ nor $(y_n)$ converged at all? Their sum still might. $\endgroup$ – user98602 May 12 '15 at 5:53
  • $\begingroup$ oh thank you, I find a counterexample that if the sum of two sequences $x_n+y_n$ if convergent, then $x_n$ and $y_n$ don't have to converge $\endgroup$ – topos May 12 '15 at 5:58
  • $\begingroup$ In the proof of "(1) sum of two open sets is open", how can you say that $B_\eta(x+y)$ is in X+Y $\endgroup$ – Ajit Mar 21 '17 at 6:07
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Something really useful to learn for these type of exercises is the following piece of information.

Let $y\in \mathbb{R}^{n}$ be fixed and define $\varphi_{y}:\mathbb{R}^{n}\to \mathbb{R}^{n}$ by setting $\varphi_{y}(x)=x+y$. Now $\varphi_{-y}\circ \varphi_{y}(x)=(x+y)-y=x$ and similarly $\varphi_{y}\circ \varphi_{-y}(x)=x$. Hence $\varphi_{y}$ is a bijection and $\varphi_{y}^{-1}=\varphi_{-y}$. Both $\varphi_{y}$ and $\varphi_{-y}$ are continuous functions so $\varphi_{y}$ is a homeomorphism.

Using the above property, many exercises of this sort become much easier.

  1. In particular, for any open set $X\subseteq \mathbb{R}^{n}$ the set $X+y$ is an open set for all $y\in\mathbb{R}^{n}$ (since $\varphi_{y}$ is a homeomorphism). In particular, \begin{align*} X+Y=\bigcup_{y\in Y}(X+y) \end{align*} is an open set as the union of open sets.

  2. Since arbitrary unions of closed sets aren't closed, we can't apply the above reasoning. But there is a standard counter example for this part. Take $X=\{-n+\frac{1}{n}:n\in\mathbb{N}\}$ and $Y=\mathbb{N}$. Both $X$ and $Y$ are closed but $X+Y=\left\{(m-n)+\frac1n\:;\:m,n\in\mathbb{N}\right\}$ has a subset $A:=\{\frac{1}{n}:n\in\mathbb{N}\}$, which has a limit point at zero but $0\notin X+Y$. Hence $X+Y$ is not closed.

  3. Since $X$ and $Y$ are compact then $X\times Y\subseteq\mathbb{R}^{2n}$ is compact, and use the continuity of the function $(x,y)\mapsto x+y$ and the fact that continuous images of compact sets are compact, to conclude that $X+Y$ is compact.

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    $\begingroup$ It's not true that $X+Y = \{\frac 1 n: n\in \mathbb N\}$, because, for example, $1+\frac 1 n = 2 + (-1+\frac 1 n) \in X+Y$. But $\{\frac 1 n: n\in \mathbb N\}$ is a subset of $X+Y$, which still yields a counterexample. $\endgroup$ – Jack Lee May 12 '15 at 14:41
  • $\begingroup$ @JackLee: Right, good call. I totally overlooked that one. Thanks for pointing it out, I'll edit it. $\endgroup$ – T. Eskin May 12 '15 at 14:43
  • $\begingroup$ Can these be applied in any NLS?@T.Eskin $\endgroup$ – sani Apr 9 '18 at 6:56
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3)The sum is a continuous operation. The image $X+Y$ of the compact set $X×Y$ is therefore compact

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  • $\begingroup$ Can these be applied in any NLS?@Arpit Kansal $\endgroup$ – sani Apr 9 '18 at 6:56
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A famous counterexample is the sum of the ice-cream cone $$ A=\{x\in \mathbb R^3: x_2^2 + (x_3-x_1)^2 \le x_1^2, \ x_1\ge 0\} $$ and the line $$ B=\{x\in \mathbb R^3: x_2=x_3=0\}. $$ Then $$ A+B = \{ x\in \mathbb R^3: \ x_3>0 \text{ or } x_2=x_3=0\}, $$ which is not closed: $x\in A+B$ means there is $t\in \mathbb R$ such that $x- \pmatrix{t\\0\\0}\in A$, that is, $x_1-t\ge0$ and $$ x_2^2 + (x_3-(x_1-t))^2 \le (x_1-t)^2, $$ which is equivalent to $$ 0\le (x_1-t)^2 - (x_3-(x_1-t))^2- x_2^2 = 2x_3(x_1-t)- x_3^2 - x_2^2 . $$ Hence the point $(0,1,0)$ is not in $A+B$, but $$ \pmatrix{ 0\\ 1 \\ 1/n} = \pmatrix{n+1\\1\\1/n} + \pmatrix{-(n+1)\\0\\0} \in A+B. $$


As you might already know, the sume $A+B$ can be proven to be closed if $A$ is compact and $B$ is closed.

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The set of integers,Z and qZ where q is a rational number, are closed subset of R but Z+qZ is not closed in R as it is dense in R.

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  • $\begingroup$ I think that $q$ should be irrational. $\endgroup$ – Watson Apr 3 '16 at 16:37

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