Why are $M$ and $N$ closed but $M+N$ is not closed

Question

I read that if $M$ and $N$ are closed subsets of a normed space, $M+N$ need not be closed.

For instance, given a continuous function $f : X \rightarrow Y$, let $M := \{(x,f(x)) : x\in X \}$, and $N:= \{ (x,0):x \in X \}$.

Now, if we set $f:= e^x$, the text says that $M+N$ is open in the space $X \times Y$.

I do not understand,... why are $M$ and $N$ closed subsets of $X \times Y$?, but $M+N$ is open?

Answer 1

$M= \{(x,f(x)) |x\in X\}$. And $N=\{(x,0)|x\in X\}$.

So $M + N =\{(x+w,f(x))|x \in X, w\in X\}$.

If $f(x) = e^x$ and $X=Y =\mathbb R$.

$M = \{(x,e^x)|x\in \mathbb R\}=$ the graph of $y=e^x$. This closed because in is a single no width graph.

$N = \{(x,0)|x\in \mathbb R\}=$ the x-axis, a straight one-dimensional line.

$M+N = \{(x+w, e^x)| x,w \in \mathbb R\}=$ ?????

No for any $(\omega, \zeta)$ we can get $\zeta = e^x$ if i) $\zeta> 0$ and $x =\ln \zeta$. And we can get $x+w = \omega$ by letting $x =\ln \zeta$ and $w = \omega - \ln \zeta$).

So $M+N = \{(\omega, \zeta)|\omega , \zeta \in \mathbb R; \zeta > 0\}=$ the positive half plane.

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$M+N= \{m+n|m\in N, n\in N\}$ is not a geometric construction and the values of $m_k$ that are added to a value of $n_j$ don't have to have anything in common and all sums must be allowed. So $(w, 0) + (x,e^x) = (w+x, e^x)$ although the $x$ must be tied to the $x$ in $e^x$ the $w$ can be anything at all.

Answer 2

In your example $M+N=\{(t,s): t \in \mathbb R, s>0\}$ and Hence it is open: to see this note that any poin tin $M+N$ is of the type $(x,0)+(y,e^{y})$ which is certainly of the type $(t,s)$ with $t \in \mathbb R, s>0$. Conversely let $t \in \mathbb R, s>0$ and write $(t,s)$ as $(t-y,0)+(y, e^{y})$ where $y =\ln s$. So $(t,s) \in M+N$.

Answer 3

So here, $X = Y = ℝ$ and $f = \exp$.

• $M = \operatorname{graph} f$ is closed because it’s the zero set of the expression $y - f(x)$. You can also employ the closed graph theorem if you are into sledgehammers.

• $N = \operatorname{graph} 0$ follows the same logic – it’s the zero set of the expression $y$.

• But $M + N = ℝ × \operatorname{img} f = ℝ × (0..∞)$.

You can see the latter by noting that $N$ is the horizontal line going through the origin and so $M + N = \bigcup_{m ∈ M} m + N$ is the union of horizontal lines going through points $m ∈ M$ with $M = \operatorname{graph} f = \operatorname{graph} \exp$. Because $\exp$ surjectively hits $(0..∞)$, the identity follows.