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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

Then

  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

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

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