Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to prove that $(\mathbb{N},\tau)$ is a topological space, where $\tau=\{ \mathbb N,\emptyset\}\cup\{\{1,\dots,n\}:n\in \mathbb{N}\}$.

I need a hint to prove that an arbitrary union of elements of $\tau$ is in $\tau$.

share|improve this question
add comment

3 Answers 3

Hint: either the union is infinite or it has a largest element.

share|improve this answer
add comment

Your hint is: consider the maximum number in the family over which you are taking the union.

share|improve this answer
add comment

Let $\{U_i: i\in I\}$ be an indexed family of elements of $\tau$. Then $\bigcup_{i\in I}{U_i}=\mathbb N$, or $\bigcup_{i\in I}{U_i}=\emptyset$, or $\bigcup_{i\in I}{U_i}=U_k$ for some $k\in I$ such that $\max{U_k}\ge \max{U_j}$ for all $j\neq k$. In all cases, the union is in $\tau$.

Is this correct?

share|improve this answer
    
That's right, because if infinitely many of the finite intervals are in your family then the union is $\mathbb{N}$, otherwise there is a largest finite interval as you say. –  Trevor Wilson Mar 1 '13 at 3:46
    
If $\mathbb N$ is in the union then it's infinite as well. The intersection in the other hand cannot be infinite, right? –  saadtaame Mar 1 '13 at 3:48
    
Yes, and yes (unless you intersect the family consisting only of $\mathbb{N}$.) –  Trevor Wilson Mar 1 '13 at 3:52
    
You don't have to be so vague about $k$. But you need a better notation. Let's say write the set $\{1,\ldots,n\}$ as $[n]$. Then for example we can say that $[a]\cup[b] = [\max(a,b)]$ for example. Then each $U_i$ in your union is $\left[n_i\right]$ for some natural number $n_i$, and $\bigcup_{i\in I} U_i = \bigcup_{i\in I} \left[n_i\right] = \left[{\max_{i\in I} n_i}\right]$, this maximum exists, or the union is $\mathbb N$ if there is no maximum. (Ignoring special cases like all the $U_i$ being empty, $I$ being empty, etc.) –  MJD Mar 1 '13 at 4:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.