Connectedness and fineness of topology

[radou]

Connectedness and "fineness" of topology

Homework Statement

Let T and T' be two topologies on X, with T' finer than T. What does connectedness of X in one topology imply about connectedness in the other?

The Attempt at a Solution

Assume (X, T) is connected, so there don't exist two disjoint, open and non-empty sets U, V whose union is X. Open sets U and V in T can be written as unions of basis elements Bx and By, where x are elements from U and y from V. Since T' is finer than T, U and V can as well be written as unions of basis elements B'x and B'y from T', where x and y are elements in U and V, respectively (since for any element x of U and Bx of T containing x there exists a basis element B'x of T' which contains x and is contained in Bx). So, connectedness of X in T implies connectedness of X in T', right?

Thanks for any replies, I hope I was clear enough, too lazy to TeX. :)

 

[radou]

Btw, another question, of logical nature.

It is almost obvious that if (X, T) is not connected, neither is (X, T'), because of a similar argument, as the upper one. Does this mean that the negation of this statement is automatically true? It may be a stupid question, but I'm curious.

 

[radou]

Any thoughts? I could make pertty much use of this fact, for example, I'm just solving the problem which asks to check whether R∞ is connected in the uniform topology. I know it's connected in the product topology, and since the uniform topology is finer than the product topology, if the above fact was true, I could conclude that it is connected in the uniform topology, too.

 

[ystael]

The reason for your confusion in the second post is that your argument in the first post proves the opposite of what you said it proves. Suppose $$(U, V)$$ disconnect $$(X, T)$$, that is $$U, V \in T$$ are a witness to the disconnectedness of $$(X, T)$$ because they are disjoint and their union is $$X$$. Because $$T'$$ is finer than $$T$$, that is $$T \subset T'$$, also $$U, V \in T'$$; therefore $$(U, V)$$ disconnect $$(X, T')$$. That is, if $$(X, T)$$ is not connected, neither is $$(X, T')$$. (Basis elements are irrelevant and not necessary to this argument.)

 

[radou]

That is, if $$(X, T)$$ is not connected, neither is $$(X, T')$$. (Basis elements are irrelevant and not necessary to this argument.)

OK, this is clear to me now.

I'll quote a post from another thread I just found:

The "contrapositive" of the statement "if P then Q" is "if not Q then not P". Notice that we have not only changed each part to "not", we have swapped hypotheses and conclusion.

If the hypotheses cannot be true when the conclusion is false then knowing that the hypothesis is true tells us that the conclusion is true. A statement is true if and only if its contrapositive is true.

So, if (X, T') is connected, so is (X, T), right?

 

[ystael]

That's correct.

 

[radou]

OK, thanks.