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For any subset A of a given metric space, closure of A is

A) open B) closed C) neither open nor closed D) none

I think that none of the above options are correct. It should be either open or closed

RKR
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Heet Modi
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  • First, what is your definition of closure? It is called "closure" for a reason :) Second, I am not sure if your last sentence is implying that you think sets are either open or closed. In any case this is not true: there are sets that are both open and closed, as well as sets that are neither open nor closed. – angryavian Jan 20 '17 at 05:58
  • Closure of a set is Union of a set and it's derived set(set of limit points) – Heet Modi Jan 20 '17 at 05:59
  • Okay, and what is the definition of closed? One of those answers is very unambiguously correct. – fleablood Jan 20 '17 at 08:01
  • It's a matter of proving that A and the closure of A have the same limit points. That should be a proposition in your text. – fleablood Jan 20 '17 at 08:27
  • Consider R the real numbers. [0,1] is closed. (0,1) is open. [0,1) is neither open nor closed. Both R and the empty set are both open and closed. – fleablood Jan 20 '17 at 08:31

2 Answers2

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The closure of a set $A$, by definition, is the smallest closed subset of the metric space (or topological space) which contains $A$. So the closure of $A$ is a closed set.

RKR
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  • sinx= 0.5 where x belongs to R is a set whose closure is open as derived set is null set and set itself is open – Heet Modi Jan 20 '17 at 06:03
  • So? Just because it is open doesn't mean it isn't closed. The empty set is both open and closed. So the closure is still closed. – fleablood Jan 20 '17 at 08:10
  • The definition of closure depends on your text. The op's text definition is that it is the union of the set with its limit points. It's a theorem that it is closed and any subset containing A will not be closed. But you are correct that it is basic that the closure is closed. – fleablood Jan 20 '17 at 08:14
  • Um, the set x so that sin x =.5 is not open. It is closed. It has no limit points. So it's closure is itself. But it is closed. – fleablood Jan 20 '17 at 08:33
  • If a set has no limit points, as your set doesn't, then the set is closed. There are no limit points for the set to contain so the set contains all the limit points there are. All zero of them. – fleablood Jan 20 '17 at 08:36
  • @fleablood Out of curiosity, how did you figure out what the OP's text definition is? What text is the OP using? – bof Jan 20 '17 at 10:02
  • The OP said so in a comment: "closure of a set is Union of a set and it's derived set(set of limit points) ". I have no friggin' idea which text she is using. – fleablood Jan 20 '17 at 19:25
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Let $(X,d)$ be the metric space with metric $d$.

Let $A \subset X$ be any subset of $X$.

Then closure is defined as union of $A$ and it's limit points.

This means closure of $A$ contains all it's limit points in $X$.

But according to the definition of closed sets, we know that the closed sets are those who contain all their limit points.

So we conclude that closure of $A$ is a closed set in $X$.

EDIT : Lemma : Limit points of $A$ and $cl(A)$ are same.

proof : We show that $D(A) \subset D(cl(A))$ and $D(cl(A)) \subset D(A)$. This will imply $D(A)=D(cl(A))$

Let $y \in D(A)$. Then there exists a sequence $\{y_n\}$ in $A$ such that $y_n \to y$. Since $cl(A)=A \cup D(A)$, we see $y \in cl(A)$ as well as $y_n \in cl(A) \; \forall n.$ Hence $y$ is a limit point of $cl(A)$. That is $y \in D(cl(A)).$ This gives $D(A) \subset D(cl(A))$.

Now let $y \in D(cl(A))$. Then there exists a sequence $\{y_n\}$ in $cl(A)$ such that $y_n \to y$. We have $y_n \in cl(A)=A \cup D(A) \; \forall n.$ If infinitely many $y_n$ are in $A$ as well as $D(A)$. Then we can consider subsequence $y_{n_k}$ of those in $A$. So $y \in D(A)$. If all terms of $y_n$ are in $D(A)$, then $y \in D(D(A))$. It is an excercise to show that $D(D(A)) \subset D(A) \Rightarrow y \in D(A)$.

This gives $D(cl(A)) \subset D(A).$

Hence completes the proof.

Note - I provided proof for the fourth line in my explanation above as suggested by @fleablood in the comments.

Error 404
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  • Consider rhe set sinx= 0.5 where x belongs to R is a set whose closure is open as derived set is null set and set itself is open – Heet Modi Jan 20 '17 at 06:05
  • I did not understand what you said. Do you mean the set ${x \in \Bbb R : \sin x = \frac 12 }$ is open? – Error 404 Jan 20 '17 at 06:07
  • Yes.Correct me if I am wrong – Heet Modi Jan 20 '17 at 06:09
  • No problem. This set is actually closed in $\Bbb R$. See that $x=n \pi + (-1)^n \frac {\pi}6 ; n \in \Bbb Z$. – Error 404 Jan 20 '17 at 06:18
  • ${x \in \mathbb R : \sin x = \frac{1}{2}} = \sin ^{-1}({\frac{1}{2}})$, which is the inverse image of the closed set ${ \frac{1}{2}}$ under the continuous function, so it is closed. – RKR Jan 20 '17 at 06:22
  • Yes @HoneyBee. :) – Error 404 Jan 20 '17 at 06:23
  • Another proof : $cl(A)=A \cup D(A)$ where $cl(A)$=closure of $A$ and $D(A)$= derived set of $A$. SInce $D(A)=\emptyset$. So $cl(A)=A \cup \emptyset=A$. Hence $A$ is closed. – Error 404 Jan 20 '17 at 06:27
  • You logic is faulty. X contains all the limit points of A. Not X. Suppose X has limit points that are not limits points of A. Then those limit points will not nescessarily be in the union of A and the limit points of A. So X need not have the limit points of X. There is a explanation as to why this never happens but you did not address it in your answer. – fleablood Jan 20 '17 at 08:06
  • @heet modi. The empty set is open and closed. So the closure is still closed. – fleablood Jan 20 '17 at 08:08
  • @fleablood I did not understand your argument. limit points of $A$ and limit points of $cl(A)$ are same right? – Error 404 Jan 20 '17 at 08:15
  • They are, but why? That is something that must be proved. – fleablood Jan 20 '17 at 08:16
  • @fleablood Okay I'll edit my answer then. With the proof. – Error 404 Jan 20 '17 at 08:17
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    Hint. If $d $ is limit point of $X$ then every neighborhood of $d$ has a point of $b\in X$. If that point isn't a point of $A $ then $b$ is a limit point of $A $. So find a neighborhood of $b$ that is completely inside the neighborhood of $d $ (you may have to argue why that can always be done). It, the neighborhood around $b $ will have a point in $A $ so that point is in the neighborhood of $d $. So d is a limit point of A as well as X. So X and A have the same limit points. – fleablood Jan 20 '17 at 08:26
  • Let us continue this discussion in chat. – Error 404 Jan 20 '17 at 09:29