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
Best Answer
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.
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.
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.
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.