We have $X = \mathbb N$. The topology is generated by the basic sets $A_n = \{n,n+1,n+2,...\} , n\in \mathbb N$. Is $X$ connected and compact? My guess is that if we take any two open sets, one of them has to be contained in the other and hence, $X$ is both connected and compact.

  • $\begingroup$ Your conclusion is right. $\endgroup$ Jan 24 at 9:28
  • $\begingroup$ Hint: If $S$ is any nonempty set of natural numbers and $N$ is the smallest element of $S$ then $\bigcup_{n\in S}A_n=A_N$. $\endgroup$ Jan 24 at 10:23

1 Answer 1


Yes, correct !

Formally, the closed sets are of the form $\{1,2\dots,n\}$. So the only clopen sets are empty and whole set itself, so connected.

Compactness follows from the fact that $A_1=\Bbb N$ is a member of every open cover of $\Bbb N$ (as the only open set that contains $1$).

  • $\begingroup$ yes..edited.. Thanks $\endgroup$ Jan 24 at 10:27

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