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I wanted to show that $(0,\infty ) $ is connected in $R_k$ topology.

OPen set in $R_k$ is $(a,b)$ or $(a,b)\setminus K $where $K=\{1/n \mid n\in \mathbb N\}$

As open interval In $(1,\infty ) $ in R and k topology same so connected On the contrary if not

$(0,1) = C\cup B$ where $C$ and $B$ are open

How to arrive at the contradiction that I don't get

Please, can anyone give me a hint so that I can complete this?

ANy Help will be appreciated

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  • $\begingroup$ What do you mean by $(a, b) K$ here? $\endgroup$ – Theo Bendit Apr 4 at 0:23
  • $\begingroup$ Are you saying that these are the open sets of your topology or that they're a basis for the topology? Either way I see a problem. If they are the open sets, I don't think you've defined a topology at all because the union of two disjoint open intervals is not open. If they are a basis, then I don't see what $(a, b) \setminus K$ adds because all sets of this form are a union of open intervals. $\endgroup$ – Robert Shore Apr 4 at 1:02
  • $\begingroup$ Sorry Sir for that . Actually I wanted to define K-topology as it is generated by (a,b) and (a,b)\K $\endgroup$ – MathLover Apr 4 at 1:17
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On $(0, \infty)$ the usual topology and that of $\mathbb{R}_K$ coincide, as $K$ is closed in $(0,\infty)$ in both subspace topologies. As it is connected in the usual one, it's connected in both.

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