# prove or disprove on topological space the following:

prove or disprove on topological space

a). Let (X,T ) be a topological space and suppose U⊆ X. Then Int(Ext(U)) ⊆ Ext(Int(U)).

b). Let (X,T ) be a topological space and suppose U⊆ X. Then Ext(Int(U)) ⊆ Int(Ext(U)).

where Ext is the exterior of the subset and Int is the interior of subset

for part b

counter-example, I find on my book.

Let X=R , T ={U⊆R:1∈U or U= empty set}

U={1,2}

so , Int U= empty set , Ext (U)=R-{1,2}

Ext(Int(U))=R and Int(Ext(U))=R-{1,2}

Hence, Ext(Int(U)) ⊈ Int(Ext(U)).

is it correct my answer for part b ?

• What do you call Ext(U)? – Régis May 6 '17 at 20:10
• It calls exterior – dr.rise May 6 '17 at 20:11
• No, for part a you cannot claim int U = U. That is the same as assuming U is open. – William Elliot May 6 '17 at 20:29
• No, for part b with a discrete topology, int {0,1} is not empty. – William Elliot May 6 '17 at 20:33
• $\operatorname{Ext}$ reverses inclusions: $A \subseteq B$ implies $\operatorname{Cl}(A) \subseteq \operatorname{Cl}(B)$ so $\operatorname{Ext}(B) \subseteq \operatorname{Ext}(A)$. – Henno Brandsma May 6 '17 at 22:19

A more extreme but more familiar example for b): let $U = \mathbb{Q}$ in $X =\mathbb{R}$ in the usual topology.

Then $$\operatorname{Int}(U) = \emptyset \text{ so } \operatorname{Ext}(\operatorname{Int}(U)) = \operatorname{Ext}(\emptyset) = \mathbb{R}$$ while

$$\operatorname{Ext}(U) = \emptyset \text{ so } \operatorname{Int}(\operatorname{Ext}(U)) = \emptyset$$

$$\operatorname{Int}(U) \subseteq U \text{ so } \operatorname{Ext}(U) \subseteq \operatorname{Ext}(\operatorname{Int}(U))$$ and as the left hand set is open, it equals its interior so indeed
$$\operatorname{Int}(\operatorname{Ext}(U)) = \operatorname{Ext}(U) \subseteq \operatorname{Ext}(\operatorname{Int}(U))$$