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Prove that the following collection of subsets defines a topology on the set of natural numbers $\Bbb N$:

$\emptyset,\Bbb N,U_n=\{1,2,\dots n\},n\in \Bbb N$.

Is $\Bbb N$ compact in this topology? What are the continuous functions from this space to the space \Bbb R$ of real numbers with standard topology?

Attempt:

$1.\emptyset,\Bbb N$ are in the collection.

$2.$ Since $U_n$'s are in the collection ,to show that $\cup_{n=1}^\infty U_n$ is also in the collection.

Now $\cup_{n=1}^\infty U_n=U_m$ for some $m$ where $m$ is the largest element (if the union is finite) and $\cup_{n=1}^\infty U_n=\Bbb N$ if the union is infinite.

$3.$ Since $U_n$'s are in the collection ,to show that $\cap_{n=1}^k U_n=U_p$ is also in the collection where $p$ is the smallest of all the elements in the $U_n$.

$\Bbb N$ is not compact since $\Bbb N=\cup _{n=1}^\infty U_n$ where $U_i's$ are open which does not have a finite subcover because if $\Bbb N=\cup _{n=1}^k U_n$ then if I take the maximum element from the $U_n$ say $p$ then $p+1\notin \cup _{n=1}^\infty U_n$.

But I am unable to find the continuous functions?Will you please give some hints?

Also please check if the arguments above are correct/not?

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    For compactness I suggest, just to make it easier to write and read, to take $p = k$, as $\forall n \in \Bbb N, U_n \subset U_{n+1}$, so this way you can use $k+1\in\Bbb N$ but $k+1 \notin \bigcup_{i=1}^k U_i$ – 2017-01-15
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    Why the downvote?? – 2017-01-15
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    I didn't downvote; by the way, until now the only continuous functions I could find are constants – 2017-01-15

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The proof that it is a topology is fine, as are the compactness remarks.

Note that any two non-empty open sets intersect non-emptily.

So suppose a continuous function $f$ into a Hausdorff $X$ has two values $f(p) \neq f(q)$, for some $p,q$. If $f(p)$ and $f(q)$ have disjoint open neighbourhoods $U$ and $V$, then $p \in f^{-1}[U]$ and $q \in f^{-1}[V]$ are disjoint non-empty and open in $\mathbb{N}$. This cannot be.