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How to find closed subsets of $A,B$ such that $\lambda A+(1-\lambda)B$ is not closed for a fixed $0<\lambda<1$.

As is well-known that if $A,B$ are bounded closed subsets of $\Bbb R^n$, then for any $0<\lambda<1$, $\lambda A+(1-\lambda)B$ is bounded and closed.

I'm suspecting that if $A,B$ at least one is unbounded, then $\lambda A+(1-\lambda)B$ need not to be closed. But I have no idea.

$A$ the $x$ axis, $B$ the $y$ axis, then $\lambda A+(1-\lambda)B$ is the whole $\Bbb R^2$, since $(x,y)=\lambda(x/\lambda,0)+(1-\lambda)(0,y/(1-\lambda))$. It is closed...

xldd
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  • Are you trying to prove that it won't be closed? Because the conclusion that it will be bounded and closed, surely fails. – Idividedbyzero Jun 03 '24 at 07:39
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    @oliH. bounded part should be trivial, so likely the OP had an impression that "bounded" condition is crucial for the closeness of the result – SBF Jun 03 '24 at 07:43
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    $\frac 1 2(2\mathbb Z)+\frac 1 2 (2\sqrt 2\mathbb Z)$ is (countable and) dense, hence not closed. – Kavi Rama Murthy Jun 03 '24 at 07:45
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    The graph of $\frac1x$, ${(x,1/x):0\neq x\in\mathbb{R}}$ and the y-axis $Y={0}\times \mathbb{R}$ in $\mathbb{R}^2$, gives the open set $\mathbb{R}^2\setminus Y.$ – Idividedbyzero Jun 03 '24 at 07:51
  • See https://math.stackexchange.com/q/124130/42969 – I suggest to close this as a duplicate. – Martin R Jun 03 '24 at 08:13

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Let $A = \{(t, \frac{1}{t}) \mid t > 0 \}$ and $B = \{0\} \times \mathbb{R}$. For every $\lambda \in (0, 1)$ we have $$ C_\lambda = (1 -\lambda)A +\lambda B = (0, +\infty) \times \mathbb{R}. $$ The sets $A$ and $B$ are closed, but $C_\lambda$ is not (at least for $\lambda \in (0, 1)$).

Daniel N
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