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A set $M \subseteq\mathbb R$ is called open if $R \setminus M$ is closed. Show that for a set $M \subseteq \mathbb R$ holds: $$M\text{ open}\iff \forall x\in M \exists \epsilon>0 : \{y\in\mathbb R\mid|x-y|<\epsilon\}\subseteq M$$

Approach: Well the direction => would be: If M is open, then IR \ M is closed, i.e. in every convergent sequence of complement in IR\M isr also the limit in IR \ M but at this point I do not get further

Thomas Andrews
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xx33xx44
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  • For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User May 29 '23 at 15:29
  • This direction is perhaps easier by contradiction. – nasekatnasushi May 29 '23 at 15:30
  • Hint: if $M$ does not satisfy the condition “$\exists\epsilon>0…$” then you can construct a convergent sequence in $\Bbb R\setminus M$ which has its limit in $M$ – FShrike May 29 '23 at 15:30
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    But also questions about the ‘sequential characterisation of closed sets’ are common, this is probably a duplicate – FShrike May 29 '23 at 15:31
  • There are a couple of possible definitions 9f "closed." What definition are you us8ng? This is all about the definitions, and the first step is to write out what it means for $\mathbb R\setminus M$ to be closed. – Thomas Andrews May 29 '23 at 15:37
  • So our definition is: A set A is called closed if for every convergent sequence in A its limit is also in A. – xx33xx44 May 29 '23 at 15:39
  • And to show it with contradiction it would say: yn ∈ {y ∈ R | |x - y| < 1/n} for all n ∈ N Since for each n there exists a y ∈ R \ M lying in {y ∈ R | |x - y| < 1/n}, for each n we can choose a yn ∈ R \ M lying in {y ∈ R | |x - y| < 1/n}. Since the sequence (yn) lies in R \ M , it has a limit y ∈ R \ M. Since R \ M is closed, the limit y lies in R \ M . Since y lies in R \ M but M is open, the limit y must lie in M, which is a contradiction. Does this make sense – xx33xx44 May 29 '23 at 15:40
  • Related. – Anne Bauval May 29 '23 at 16:28
  • Your title is terrible. Definitions are not proved. – jjagmath May 29 '23 at 17:19

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I couldn't find a duplicate, so here is a short answer.

First notice that your definition of "$M$ is open" as "whenever $x_n\to x,$ if $\forall n\quad x_n\notin M$ then $x\notin M$" rewrites: "whenever $x_n\to x,$ if $x\in M$ then $\exists n\quad x_n\in M$".

The node of the problem is therefore to prove, given $x\in M,$ that

$(1)$ for every sequence $(x_n)$ converging to $x,$ at least one $x_n$ belongs to $M$

is equivalent to

$(2)$ $\exists\epsilon>0\quad(x-\epsilon,x+\epsilon)\subset M.$

By definition of the convergence of sequences, $(2)\implies(1).$

Conversely, let us prove $(\neg2)\implies(\neg1).$ Assume $\forall n\in\Bbb N\quad(x-1/n,x+1/n)\not\subset M.$ Picking for each $n$ some $x_n\in(x-1/n,x+1/n)\setminus M$ gives a sequence $(x_n)$ contradicting $(1).$

Anne Bauval
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  • Is that the case in both sides so from M is open than we see … and the other side from the definition we can say that M is open – xx33xx44 May 29 '23 at 17:39
  • Excuse me, I don't understand the meaning (not even the syntax) of your sentence. Is it a question? (or several?) Can you please reformulate it? (or them?) – Anne Bauval May 29 '23 at 18:27
  • I mean with <=> it means that we must show the definition in both ways, so in the way => and the way <=. So in the case of => If M is open and we can conclude that x∈M ∃ε>0 : {y∈R||x−y|<ε}⊆M and in the other way <= if we have x∈M ∃ε>0 : {y∈R||x−y|<ε}⊆M than we can include that M is open – xx33xx44 May 29 '23 at 18:54
  • What does "show the definition" mean? What do you put on each side of your <=>? Are you talking about the second sentence of my answer? ("First notice that your definition [...] rewrites [...]") Here, "rewrites" means "is clearly equivalent to". – Anne Bauval May 29 '23 at 18:56