Closure of Intersection may not equal Intersection of Closures/Examples/Arbitrary Subsets of Real Numbers

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Examples of Closure of Intersection may not equal Intersection of Closures

Let $H$ and $K$ be subsets of the set of real numbers $\R$ defined as:

\(\ds H\) \(=\) \(\ds \openint 0 2 \cup \openint 3 4\)
\(\ds K\) \(=\) \(\ds \openint 1 3\)


Let $\map \cl H$ denote the closure of $H$.


Then:

$H \cap \map \cl K$
$\map \cl H \cap K$
$\map \cl H \cap \map \cl K$
$\map \cl {H \cap K}$

are all different.


Proof

From Closure of Open Real Interval is Closed Real Interval:

\(\ds \map \cl H\) \(=\) \(\ds \closedint 0 2 \cup \closedint 3 4\)
\(\ds \map \cl K\) \(=\) \(\ds \closedint 1 3\)


Hence by definition of set intersection:

\(\ds H \cap \map \cl K\) \(=\) \(\ds \paren {\openint 0 2 \cup \openint 3 4} \cap \closedint 1 3\) \(\ds = \hointr 1 2\)
\(\ds \map \cl H \cap K\) \(=\) \(\ds \paren {\closedint 0 2 \cup \closedint 3 4} \cap \openint 1 3\) \(\ds = \hointl 1 2\)
\(\ds \map \cl H \cap \map \cl K\) \(=\) \(\ds \paren {\closedint 0 2 \cup \closedint 3 4} \cap \closedint 1 3\) \(\ds = \closedint 1 2 \cup \set 3\)
\(\ds \map \cl {H \cap K}\) \(=\) \(\ds \map \cl {\openint 0 2 \cup \openint 3 4} \cap \openint 1 3\)
\(\ds \) \(=\) \(\ds \map \cl {\openint 1 2}\)
\(\ds \) \(=\) \(\ds \closedint 1 2\)

All defined sets, as can be seen, are different.

$\blacksquare$


Sources

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