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Does closure of a set mean, only adding boundary values if the set is open and leave it as it is if the set is closed?

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Don’t forget, there are sets that are neither open nor closed. An example is $\{z\in\Bbb C:1\le|z|<2\}$. To get the closure of such a set, you must include its boundary points, even though it’s not an open set. –  Brian M. Scott Nov 5 '12 at 1:34

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The best way to think of closure of a set $E$ is as the smallest closed subset of the space containing $E$. Such a thing exists because any intersection of closed sets is closed; just intersect all closed subsets of the space (remember: the whole space itself is closed) containing $E$.

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Generally, and somewhat loosely, closure of a set means add whatever is needed, and only that which is needed, so that the new set has a certain property. Your question indicates that the property you have in mind is being topologically closed yet the tags you used indicate that you are interested in the more general case. So, yes, if you are interested in closure in the topological meaning then the closure of a set $A$ is the smallest possible super-set $B$ which is itself closed. Loosely speaking, $B$ is obtained from $A$ by adding 'boundary values' as you say but you have to be very precise about what you mean by 'boundary value'.

Other examples include: The closure of a relation $R$ on a set $A$ with respect to the property of being transitive is usually called the transitive closure of $R$ (it is obtained by adding just those elements needed to make $R$ transitive). The closure of a subset $S$ of a group $G$ under the group operation is more commonly known as the subgroup generated by $S$. Many more examples exist in practically all realms of mathematics.

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Let $A'$ be the set of all limit points of $A$. Then the closure of $A$ is equal to $A\cup A'$. To answer your question, no. A point can be a limit point without being a boundary point. For instance, every interior point of $A$ is a limit point of $A$ but no interior point of $A$ is a boundary point of $A$. But yes, if the set is already closed, then yes, it already contains all of its limit points so leave it as is.

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