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Let $K$ is a non empty closed subset of $\mathbb{R}$ , then show that the set {${x+y : x\in K , y\in [1,2]}$} is closed in $\mathbb{R}$.

I know a set will be closed if it contains all it's limit points but not know how to apply this here...

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  • $\begingroup$ What is the definition of a limit point that you use? The two answers already posted are correct, one of them very detailed. Could you please indicate which part you do not follow, that needs extra attention (to justify the bounty) and additional explanations? $\endgroup$
    – Mirko
    Dec 22, 2017 at 23:56
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    $\begingroup$ This Q has appeared before. BTW it is common to write $A+B$ for $\{a+b: a\in A\land b\in B\}$ and $a+B$ for $\{a+b:b\in B\}$..... Note that the sequential compactness of $[0,1]$ in your Answers is crucial.(A sequence in $[0,1]$ has a convergent sub-sequence). For example, if $A=\Bbb N$ and $B=\{-n+\frac {1}{n+1}:n\in \Bbb N\}$, then $A$ and $B$ are closed but not compact, and $0\in \overline {A+B}\setminus (A+B)$ . $\endgroup$ Dec 22, 2017 at 23:58
  • $\begingroup$ are you aware that every bounded sequence of real numbers has a convergent subsequence? This is perhaps the main element that makes the proof work. The sum considered (in the above comment) is called Minkowski sum, there are many answers if you search, see en.wikipedia.org/wiki/Minkowski_addition and math.stackexchange.com/q/80974 and math.stackexchange.com/q/80800 there are more too math.stackexchange.com/search?q=minkowski+sum $\endgroup$
    – Mirko
    Dec 23, 2017 at 0:16
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    $\begingroup$ This question has been completely answered by gvhl. I wonder what kind of additional attention you are asking for with your bounty? $\endgroup$
    – Thomas
    Dec 23, 2017 at 10:44

2 Answers 2

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Let $(z_n)_{n \in \mathbb{N}}$ be a convergent sequence in $\{x + y: x \in K, y \in [1, 2]\}$ with limit $z$. By definition for each $z_n$ there exist $x_n \in K$ and $y_n \in [1, 2]$ such that $z_n = x_n + y_n$. Since $[1, 2]$ is compact by Heine-Borel, there must exists a convergent subsequence $(y_{n_k})_{k \in \mathbb{N}}$ such that $y_{n_k} \to y \in [1, 2]$. But then $x_{n_k} \to z - y$, i.e. we have a convergent subsequence $(x_{n_k})_{k \in \mathbb{N}}$ in $K$. But since $K$ is closed, we have that $z - y \in K$.

But now note that $z = (z - y) + y$, where $(z-y) \in K$ and $y \in [1, 2]$. That is, the limit of $(z_n)_{n \in \mathbb{N}}$ is in the form $x + y: x \in K$ and $y \in [1, 2]$, and thus $\{x + y: x \in K, y \in [1, 2]\}$ is closed.

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  • $\begingroup$ gvhi. Very nice. $\endgroup$ Dec 17, 2017 at 17:48
  • $\begingroup$ when you say "convergent sequence in" a given set, do you mean that the elements are in the set, or that the limit is also in the set. This might be a point of confusion (trying to understand why the OP put a bounty) $\endgroup$
    – Mirko
    Dec 23, 2017 at 0:21
  • $\begingroup$ @Mirko "convergent sequence in" means the first point that you mentioned, as in that the elements are a member of the set we are talking about. $\endgroup$
    – gvhl
    Dec 23, 2017 at 21:58
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Suppose $z_n$ is a sequence in $S=\{x+y: x\in K, y \in [1,2]\}$ such that $z_n\rightarrow z$. You have to show $z\in S$.

Since $ z\in S$ there are $x_n,y_n$ such that $x_n\in K$ and $ y_n\in [1,2]$ and $z_n = x_n+y_n$. Since $[1,2]$ is closed and bounded, a subsequence $y_{n_k}$ of $y_n$ converges.

Can you finish from here?

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  • $\begingroup$ sir , are you using a set $A$ is closed if every convergent sequence in $A$ converges to a limit in $A$? $\endgroup$
    – user455480
    Dec 17, 2017 at 15:12
  • $\begingroup$ @SULTAN I'm not using it, but the continuation of the proof the beginning of which I wrote down will. See the answer of gvhl. $\endgroup$
    – Thomas
    Dec 17, 2017 at 15:15

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