How does the second paragraph prove that \tau' is the same as \tau?

[camilus]

Lemma 13.2: Let X be a topological space. Suppose that C is a collection of open sets of X such that for each open set U of X and each x in U, there is an element c of C such that $x\in c\subset U$. Then C is a basis for the topology of X.

Proof: The first paragraph is trivial, it just shows that the conditions of basis are satisfied.

The second paragraph attempts to show that $\tau'$, the topology generated by C, is the same as the topology $\tau$ on X.

Can someone elaborate on the second paragraph please?

 

[micromass]

What don't you understand about the second paragraph??

You just let $\mathcal{T}^\prime$ be the topology generated by the basis $\mathcal{C}$. So by definition, a set G belongs to $\mathcal{T}^\prime$ if for every x in G, there is a set C in $\mathcal{C}$ such that

$$x\in C\subseteq G$$

So, to prove that $\mathcal{T}\subseteq\mathcal{T}^\prime$. Take G in $\mathcal{T}$. By hypothesis, there exists for every for every x in G, a set C in $\mathcal{C}$ such that

$$x\in C\subseteq G$$

So, by definition almost, we have that $\mathcal{T}\subseteq \mathcal{T}^\prime$.

The other inclusion is less obvious. So to prove that $\mathcal{T}^\prime\subseteq\mathcal{T}$ we take an element $G\in \mathcal{T}^\prime$. By hypothesis, there is for every x in G, a set C in $\mathcal{C}$ such that

$$x\in C\subseteq G$$

This implies that $G=\bigcup{C}$ is the union of all these C's. Since all the C's are in $\mathcal{T}$, it implies that G is also in $\mathcal{T}$ as union of open sets...

IS that more clear?

 

[Greg Bernhardt]

Sure, I'd be happy to elaborate on the second paragraph.

The second paragraph is essentially showing that the topology \tau' generated by the collection C is the same as the original topology \tau on X. This is important because it confirms that C is indeed a basis for the topology of X.

To do this, we first define \tau' as the collection of all unions of elements of C. This means that every open set in \tau' can be written as a union of elements from C.

Next, we show that \tau' is a topology on X. To do this, we need to show that \tau' satisfies the three axioms of a topology: (1) the empty set and X are in \tau', (2) arbitrary unions of sets in \tau' are in \tau', and (3) finite intersections of sets in \tau' are in \tau'.

First, we know that the empty set and X are in \tau' since they can be written as unions of elements from C (the empty union for the empty set, and the union of all elements in C for X).

Next, we consider an arbitrary union of sets in \tau'. This means we have a collection of sets in \tau' (which are unions of elements from C) and we want to show that their union is also in \tau'. This follows directly from the definition of \tau' as the collection of all unions of elements from C.

Finally, we consider finite intersections of sets in \tau'. This means we have a finite collection of sets in \tau' and we want to show that their intersection is also in \tau'. This can be done by noting that since each set in \tau' is a union of elements from C, their intersection is also a union of elements from C.

Now, since \tau' satisfies the three axioms of a topology, it must be a topology on X. And since every open set in \tau' can be written as a union of elements from C, we can conclude that \tau' is the same as the original topology \tau on X. Therefore, C is indeed a basis for the topology of X.

I hope that helps clarify the second paragraph for you. Let me know if you have any other questions.