Let $\mathbb Z_+ = \{1, 2, 3, .....\}$ be the set of positive integer . Let $\tau_1 := $ subspace topology on $\mathbb Z_+$ induced from the usual topology on $\mathbb R ,$

$\tau_2 :=$ order topology on $\mathbb Z_+$, i.e the topology with base

$\{ \{ x: 1 \leq x < b \} : b \in \mathbb Z+b \}$ $\cup$ $\{ \{ x: a < x < b \} : a, b \in \mathbb Z+b \}$

$\tau_3 :=$ discrete topology


  1. $\tau_1 \neq \tau_3$ and $\tau_1 = \tau_2$

  2. $\tau_1 \neq \tau_2$ and $\tau_1 = \tau_3$

  3. $\tau_1 \neq \tau_3$ and $\tau_2 = \tau_2$

  4. $\tau_1 = \tau_2 = \tau_3$

My attempt is :

Any any $a\in \mathbb Z_+$, then $(a-\epsilon , a+\epsilon)$, So $\tau_1$ is a dicrete topology

Similarly $\tau_2$ is a discrete topology

  • $\begingroup$ You are correct. $\endgroup$ – MJD Dec 23 '14 at 6:46

You are indeed correct. The argument could be written up better: for any $a \in \mathbb{Z}^{+}$, we have that $(a - \frac{1}{2}, a + \frac{1}{2})$ is open in $\mathbb{R}$, so its intersection with $\mathbb{Z}^{+}$, is open in the subspace topology $\tau_1$, and this intersection equals $\{a\}$. As all singletons are open in $\tau_1$, $\tau_1$ is the discrete topology.

$\{1\} = [1,2)$ in the order topology and for $a \in \mathbb{Z}^{+}, a \ge 2$ we can write $\{a\} = (a-1,a+1)$ (the endpoints have to lie in $\mathbb{Z}^{+}$, hence the distinction). So again $\tau_2$ is the discrete topology as all singletons are open.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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