Which one of the following is correct?a)The intersection of any finite...
Let G
1, and G
2 be two open sets. Then, if G
1 ∩ G
2 = φ, it is open
if G
1 ∩ G
2 ≠ φ, let x ε G
1 ∩ G
2implies x ε G
1, and x ε G
2implies G
1, G
2 are nbd of x.
implies G
1 ∩ G
2 is a nbd of x.
but since x is any point o f G
1 ∩ G
2 therefore G
1 ∩ G2 is a nbd o f each of its points. Hence G
1 ∩ G
2
is open.
Now, consider the open sets
implies
which is not an Open set.
implies The intersection of arbitrary number of open set need not be open. Next, let G be the union of an arbitrary family
of open sets, A being an index set. To prove that G is an open set, we shall show that for any point x ε G, it contains an open interval containing x.
Let x ε G implies
atleast one member, say
of F such that x ε
. Since,
is an set,
an open interval ln such that
Thus, the set G contains an open interval containing any point x of G. Hence, G is an open set.
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Which one of the following is correct?a)The intersection of any finite...
Explanation:
Definition: A set is said to be open if every point in the set is an interior point.
Option A states that the intersection of any finite number of open sets is open. Let's prove this statement.
Proof:
Suppose we have a finite collection of open sets: A₁, A₂, ..., Aₙ.
Let x be an arbitrary point in the intersection of these sets, i.e., x ∈ ⋂ᵢ₌₁ⁿ Aᵢ.
Since x is in the intersection, it means that x is in every set Aᵢ.
Claim: x is an interior point of the intersection of these sets.
Proof of Claim:
To show that x is an interior point, we need to find an open ball centered at x that is completely contained in the intersection of the sets.
Since x is in every set Aᵢ, there exists an open ball B(x, rᵢ) centered at x such that B(x, rᵢ) ⊆ Aᵢ for each i.
Let r be the minimum of all the radii rᵢ, i.e., r = min{r₁, r₂, ..., rₙ}.
Now consider the open ball B(x, r). We claim that B(x, r) is completely contained in the intersection of the sets.
Let y be an arbitrary point in B(x, r). We need to show that y is also in every set Aᵢ.
Since y is in the ball B(x, r), the distance between y and x is less than r, i.e., d(y, x) < />
By the triangle inequality, we have:
d(y, x) ≤ d(y, x) + d(x, y) = d(y, x) + d(y, x) < r="" +="" r="" />
Since 2r < rᵢ="" for="" each="" i,="" it="" follows="" that="" d(y,="" x)="" />< rᵢ="" for="" each="" />
Therefore, y is in every set Aᵢ, and hence, y is in the intersection of the sets.
This shows that every point x in the intersection of the finite number of open sets has an open ball centered at x that is completely contained in the intersection. Hence, the intersection is open.
Therefore, option A is correct: The intersection of any finite number of open sets is open.