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Metric Spaces: Interior of a Set

  1. Aug 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##(X,d)## be some metric space, and let ##A## be some subset of the metric space. The interior of the set ##A##, denoted as ##int A##, is defined to be ##\bigcup_{\alpha \in I} G_\alpha##, where ##G_\alpha \subseteq A## is open in ##X## for all ##\alpha \in I##. The closure of ##A##, denoted as ##\overline{A}##, is defined to be ## \bigcap_{\beta \in J} F_\beta##, where ##A \subseteq F_\beta## and ##F_\beta## is closed in ##X##.

    I am asked to show ##int A = X - (\overline{X-A})##

    2. Relevant equations


    3. The attempt at a solution

    I am first trying to show that ##int A \subseteq X - (\overline{X-A})##. Clearly if ##x \in int A##, then ##x## will also be in ##X##, because ##int A \subseteq X##. All that remains is to show that ##x \notin (\overline{X-A})##.

    I could see a direct way of showing this inclusion, so I tried to prove it by contradiction, but that did not help either. Is there a direct way of showing this? Does anyone know if a direct proof is possible? Any hints?
     
  2. jcsd
  3. Aug 17, 2016 #2
    A proof by contradiction is probably your best way forward in showing that something is *not* an element of a well defined set. Assume first that ##x## is in both ##\text{int } A## and ##\overline{X-A}##. This means, with respect to your definition of closure, that ##x## must be an element of *every* closed set that contains ##X - A##. Therefore, if we could find a closed set that contains ##X - A##, but does not contain ##x##, our contradiction would be complete.
    ##\text{int }A## is an open subset of A, so it therefore has an empty intersection with ##X - A##. What can we therefore say about the complement of ##\text{int }A## ? Do you see how this helps us in moving forward ?
     
  4. Aug 18, 2016 #3
    Yes, you hints have proved very helpful, so much so that I finished the proof. Now, though, I am having difficulty with the other inclusion. Here is what I have so far:

    Suppose that ##x \in X - (\overline{X-A})##. Then ##x \in X## but ##x \notin (\overline{X-A}) = \bigcap_{j \in J} F_j##, which means ##x \notin F_j## for some ##j \in J##. However, ##x## will be in ##X-F_j##, which is open because ##F_j## is closed.

    This is where I couldn't figure out how to proceed. I would like to show that ##X-F_j \subseteq A##, as this would mean ##X-F_j## is apart of the union ##\bigcup_{i \in I} G_i = int A##, which would conclude the proof. But I am uncertain of how to show this.
     
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