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Let $$ K \colon= \left\{\ \frac{1}{n} \ \colon \ n \in \mathbb{N} \ \right\},$$ and let the $K$-topology on $\mathbb{R}$ be the one having as basis all open intervals $(a,b)$ and all sets of the form $(a,b)-K$, where $a, b \in \mathbb{R}$ such that $a< b$.

Then this topology is finer than the usual topology on $\mathbb{R}$.

And, these two topologies are different.

But what sets are open in the $K$-topology but not in the usual topology on $\mathbb{R}$?

And, how are the subspace topologies that $(0, +\infty)$ inherits as a subspace of $\mathbb{R}$ under these two topologies, respectively, the same, as Munkres claims?

As per my understanding, any set of the form $(a,b)-K$ is again a union of one or more open intervals and is therefore in the usual topology.

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Sets of the form $(a,b)-K$ where $a<0< b$ are not open in the usual topology. This is because there is no open interval containing $0$ that does not contain an element of $K$, and given any open set $U$ in the usual topology and any point $x$ in the set there must be an open interval $I$ such that $x\in I\subset U$.

To answer the second question, the reason the subspace topology on $(0,\infty)$ is the same as the usual topology is that the intersection of $K$ with this subset is closed. All of the new open sets in the $K$ topology contain 0.

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  • $\begingroup$ absolutely. And what about the second question that I've posed? Could you please answer that too? $\endgroup$ – Saaqib Mahmood May 23 '15 at 17:23
  • $\begingroup$ Please see my edit. $\endgroup$ – Matt Samuel May 23 '15 at 23:29
  • $\begingroup$ your second paragraph is not fully clear. I do understand that the set $K$ is closed in $(0, +infty)$ in the subspace topology that $(0, +\infty)$ inherits from the $K$-topology on $\mathbb{R}$. Right? But I didn't get the rest of your argument. $\endgroup$ – Saaqib Mahmood May 24 '15 at 5:23
  • $\begingroup$ I was saying that in order for a set to be open in the K topology but not the usual topology it has to contain 0. $\endgroup$ – Matt Samuel May 24 '15 at 18:51
  • $\begingroup$ Yes, @Matt Samuel, you're right. For a set to be open in the $K$-topology but not in the usual topology, that set must contain $0$. So? How do we conclude from this fact that the subspace topologies that $(0, +\infty)$ inherits as a subspace, respectively, of the usual topology and the $K$-topology on $\mathbb{R}$ are the same? $\endgroup$ – Saaqib Mahmood Jul 6 '15 at 18:49

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