Download Quiz-2 Solutions

Point set topology (MTH-322)
(1)
(2)
(3)
(4)
Quiz-2 Solutions
To prove (1), it is enough to prove that intersection of two open dense
subsets is open, since complement of a nowhere dense set is a dense set.
Let U1 and U2 be two open dense subsets of X. To show that B := U1 ∩U2
is open and dense in X. Clearly U1 ∩ U2 is again a open set in X. To
show that B is dense, we want to show, if V is any non-empty open set in
X then V ∩ B 6= φ. As U1 is dense, U1 ∩ V 6= φ. Further U1 ∩ V is open
and thus U2 ∩ (U1 ∩ V ) 6= φ (since U2 is dense). Thus (U1 ∩ U2 ) ∩ V 6= φ.
Let X be a discrete metric space and Y be any other metric space. Let
f : X → Y be a function. Let U be a open set in Y . Then f −1 (U ) is
subset of X. So it can be written as a union of singletons, which are open
subsets in the discrete metric space. Thus f −1 (U ) is a open set being a
union of open subsets. Thus f is a continuous function.
Consider R with the lower limit topology τ 0 . This topology is given by the
basis B0 := {[a, b)|a, b ∈ R}. The standard topology τ on R is given by
the basis B := {(a, b)|a, b ∈ R}. To show that lower limit topology is finer
than the standard topology i.e τ 0 ⊃ τ . Given a basis element (a, b) for τ
and a point x ∈ (a, b) take [x, b) ∈ τ 0 which contains x and [x, b) ⊂ (a, b).
Therefore τ 0 is finer than τ . Moreover given a basis element [x, b) for τ 0 ,
there is no open interval (a, b) containing x and contained in [x, b). Thus
τ 0 is strictly finer than τ .
Let X and Y be two topological spaces. The product topology on X × Y
is the topology given by the basis
B : {U × V |U is open in X, V is open in Y }
(5) Let X be a set. The discrete topology on X is the collection of all subsets
of X. Thus the collection of all one point sets is a basis for the discrete
topology. The indiscrete topology on X is the collection consisting of only
two sets {X, φ}. Now, B := {X}, is the basis for this topology.
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