Question

Let X be a set and let T be the family of subsets U of X such that X\U (the complement of U) is at most countable, together with the empty set. a) Prove that T is a topology for X. b) Describe the convergent sequences in X with respect to this topology. Prove that if X is uncountable, then there is a subset S of X whose closure contains points that are not limits of the sequences in S.

12. be the family of subsets Ụ of X such that XU is Let X be a set and let at most countable, together with the empty set Ø. (a) Prove that J is a topology for X. (b) Describe the convergent sequences in X with respect to this topology (c) Prove that if X is uncountable, then there is a subset S of X whose closure contains points that are not limits of convergent sequences in Hint: Take S to be any proper uncountable subset of X. S.

Answer 1

$\tau =$ { $U \subset X :$ X \ U is atmost countable } .

(a) .

(i) given $\phi \in \tau$ . Also X $\in \tau$ since X\X = $\phi$ which is empty set so countable.

(ii) suppose U₁ , U₂ , ....., U_n$\in \tau$.  

$\Rightarrow$ X\U₁ , X\U₂ , ......, X\U_n are all atmost countable set.

Now , X\ ( $U_{1} \cap U_{2} \cap......\cap U_{n} ) \subseteq$ (X\$U_{1}$ ) $\cup$ ( X\(X\U_n)

U2) U..

As finite union of countable set is countable so  (X\$U_{1}$ ) $\cup$ ( X\
U2) U..
(X\U_n) is countable . Now since subset of a countable set is countable so X \ ( U₁$\cap$ U₂$\cap$ ......$\cap$ U_n ) is almost count able.

$\Rightarrow$  U₁$\cap$ U₂$\cap$ ......$\cap$ U_{n $\in \tau$​​​​​}

_{Hence $\tau$ is closed under finite intersection .}

(iii) suppose U₁ , U₂ , ....... are elements of $\tau$

$\Rightarrow$ X\U_i is a countable set for all I = 1 , 2 , 3 ,.....

Now X\ (U₁$\cup$ U₂ ......) $\subset$( X \ U₁ ) $\cup$ (X\ U₂ ) .......

As countable union of countable set is countable so X\ (U₁$\cup$ U₂ ......) Is almost countable .

$\Rightarrow$ U₁ U U₂ ....... belongs to $\tau$

Hence $\tau$ is closed under arbitrartrary union .

Hence $\tau$ os a topology on X .

(b). Let (x_n) be a sequence converges to x .

Then every open set containing x contains a point of the sequence (x_n) .

Let A ={ x₁ , x₂ , .... } is a countable set

So X\A is an open set in X contains no point of the sequence other than x .

Hence (x_n) is a eventually constant sequence of x .

I.e., only convergent sequence are eventually constant sequence .

(c). Let a $\in$ X define S = X \{a} then closure of S is {a} but a is not a limit point of any sequence in S because the only convergent sequence in X are eventually constant sequence .

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Please comment if needed.