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
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 24$