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I am confused regarding the following two sets:

  1. $[a, b]\cap(c, d)$ This is an intersection of a closed set and an open set. Is such an intersection always neither open nor closed?

  2. $[a, b] \cup [c, d)$ This is a union of a closed set and a set which is neither open nor closed. What will be the union of these two sets?

Stefan Hamcke
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Deepabali Roy
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  • Possible copy of http://math.stackexchange.com/questions/186101/example-of-neither-open-nor-closed-set?rq=1 – Matthew Levy Feb 05 '15 at 15:40
  • Note to editors: $]a,b[$ is a perfectly standard notation for an open set in many countries. To the editor who simply changed these to $[a,b]$: didn't you realise that you had destroyed the question? – TonyK Feb 05 '15 at 15:41
  • @MatthewLevy I did check the question you have mentioned. That does not explain these two cases in specific. – Deepabali Roy Feb 05 '15 at 15:45

3 Answers3

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Here are some (hopefully) helpful examples:
1. Consider the intersection of $[-2,2]$ and $(0,1)$.
2. Consider the union of $[-2,2]$ and $[0,1)$.

Sloan
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  1. The intersection of an open and a closed interval could be either open not closed, closed not open, open and closed, neither open nor closed:

Consider $$ [-1,1] \cap (-2,2) = [-1,1] \\ [-2,2] \cap (-1,1) = (-1,1) \\ [0,1] \cap (1,2) = \emptyset \\ [0,1] \cap (-\frac12, 1) = [0,1)$$

  1. The Union cannot be open but similar to 1. it can be closed or neither open nor closed.
abc
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  • In the last example you mentioned [0,1] ∩ (1/2,1) is (1/2, 1) which is open – Deepabali Roy Feb 05 '15 at 16:16
  • @DeepabaliRoy you are right, I corrected it – abc Feb 05 '15 at 16:20
  • Thanks! Just one more doubt, For the second part, how do I conclude that the set will never be open? – Deepabali Roy Feb 05 '15 at 16:26
  • @DeepabaliRoy If $a \leq c$ then no neighborhood of a can belong to the union since $a-\epsilon < a \leq c$ for all $\epsilon > 0$.

    If $c < a$ no neighborhood of c can belong to the union.

     – abc Feb 05 '15 at 16:29
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In general, we can say that the answer to 1. is no. To show that, note that if (c,d) ⊂ [a,b], then [a,b] ∩ (c,d) = (c,d), that is, an open set.

About the second one, (I believe you meant "union"), we have no means to know for sure, as no assumption has been made on, say, d ∈ [a,b]. Anyway, we can note that, if d ∈ [a,b], then the result will be a closed set, whilst in any other case the set will be neither open nor closed.

Luigi D.
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