Understand the problem

True or False? Let \(H_1, H_2, H_3,H_4 \) be four hyperplanes in \( R^3 \). The maximum possible number of connected components of \( R^3 – H_1 \cup H_2 \cup H_3 \cap H_4 \) is 14.
Source of the problem
TIFR GS 2017 Entrance Examination Paper
Topic
General Topology
Difficulty Level
Easy
Suggested Book
Topology, Second Edition, English, Paperback, by James R. Munkres

Start with hints

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If \(m\) is the number of hyperplanes and \(n\) is the dimension of the space in which the hyperplanes are intersecting then can you derive a formula which will give you the maximum possible number of connected components?
When \(m=2\) and \(n=1\) (that is, two hyperplanes are passing through a line, say) then the number of maximum possible connected components is \(2\), that is \({m \choose n} \).
When \(m=3\) and \(n=2\) (that is, three hyperplanes are intersecting in \(\Bbb R^2\) space) then the number of maximum possible connected components is \(3\), that is \({ m \choose n} \).
Hence, the maximum possible number of connected components when \(m\) hyperplanes are intersecting in \( \Bbb R^n\) is \( {m \choose 0}+{m \choose 1}+{m \choose 2}+{m \choose 3}+ \ldots +{m \choose n} \) when \( m>n\) and is \( 2^m \) when \(m<n\) and \(2^m-1\) when \(m=n+1 \).
Now, what do you think could be the maximum possible number of connected components in the given question?
The maximum possible number of connected components is \(15\). But the statement is saying \(14\). Hence the statement is false..

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