Exercise 2.2

Exercise 2.2: Let $$X = \{a,b,c\}$$ and let $$\mathcal{T} = \{\emptyset, \{a,b,c\}, \{a\}, \{b,c\}\}$$. Verify that $$\mathcal{T}$$ is a topology on $$X$$. (Hence, $$(X,\mathcal{T})$$ is a topological space.)

baruch_shahi
That $$X$$ and $$\varnothing$$ are in $$\mathcal{T}$$ is obvious.

Now, if we intersect $$\varnothing$$ with anything else in $$\mathcal{T}$$, we get $$\varnothing$$ back. Consider the following pairwise intersections $$\{a,b,c\}\cap \{a\}=\{a\}$$ $$\{a,b,c\}\cap\{b,c\}=\{b,c\}$$  $$\{a\}\cap\{b,c\}=\varnothing\,.$$ Since intersection is a commutative operation on sets, we have accounted for all possible pairwise intersections of elements of $$\mathcal{T}$$. Notice that every intersection is an element of $$\mathcal{T}$$. Thus, $$\mathcal{T}$$ is closed with respect to finite intersection.

By inspection, an arbitrary union of elements in $$\mathcal{T}$$ can never yield anything other than the elements of $$\mathcal{T}$$. More rigorously, any union containing $$\{a,b,c\}$$ must be equal to $$\{a,b,c\}$$. Any union containing $$\{b,c\}$$ but not $$\{a,b,c\}$$ must be equal to $$\{b,c\}$$. Any union containing just $$\{a\}$$ and $$\varnothing$$ is equal to $$\{a\}$$. Obviously any union only containing $$\varnothing$$ is just $$\varnothing$$. Hence, $$\mathcal{T}$$ is closed with respect to arbitrary union.

Thus, $$\mathcal{T}$$ is a topology on $$X$$.

baruch_shahi 09:21, March 22, 2011 (UTC -5)

'''Thanks for working through the details! --Steven.clontz 19:32, March 22, 2011 (UTC)'''