Initial segment topology

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 $\tau$.

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

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

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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?

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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