# How to Show $(0,\infty )$ is connected in $R_k$ topology?

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

• What do you mean by $(a, b) K$ here? – Theo Bendit Apr 4 at 0:23
• 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. – Robert Shore Apr 4 at 1:02
• Sorry Sir for that . Actually I wanted to define K-topology as it is generated by (a,b) and (a,b)\K – MathLover Apr 4 at 1:17

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.