R is disconnected with the subspace topology

[kent davidge]

I want to show that $\mathbb{R}$ is disconnected with the subspace topology. For this I considered that $\mathbb{R} = \lim_{\delta n \longrightarrow 0 } (-\infty, n] \cup [n+\delta n, \infty)$ and of course the intersection of these two open sets is empty.

What I'm not sure is about the usage of limit here. Is this ok? I personally think it's ok, because I'm using limit only for saying that $\delta n > 0$ but that $n+\delta n$ should be the closest possible real number of $n$...

 

[fresh_42]

I want to show that $\mathbb{R}$ is disconnected with the subspace topology.

Of what? Where did you find this idea? $\mathbb{R}$ is a straight line, why should it be disconnected?

For this I considered that $\mathbb{R} = \lim_{\delta n \longrightarrow 0 } (-\infty, n] \cup [n+\delta n, \infty)$ and of course the intersection of these two open sets is empty.

And doesn't cover $\mathbb{R}$, aren't open sets, and should better be something like $\bigcup_k[n+\frac{1}{k}, \ldots )$

What I'm not sure is about the usage of limit here. Is this ok?

No. It doesn't make sense. You need open sets.

 

[kent davidge]

Of what? Where did you find this idea?

I mean the subspace topology obtained by the intersection of $\mathbb{R}$ with its own usual euclidean topology. Are'nt sets like those in post #1 open in such topology? And the idea I got by wondering about.

And doesn't cover $\mathbb{R}$

I thought it did if we consider $n+ \delta n$ as the closest neighboor of $n$.

 

[kent davidge]

Or... to give the sets a better look than those in post #1... $\mathbb{R} = (-\infty,n) \cup [n, \infty)$... intersection is empty...

 

[fresh_42]

I mean the subspace topology obtained by the intersection of $\mathbb{R}$ with its own usual euclidean topology. Are'nt sets like those in post #1 open in such topology? And the idea I got by wondering about.

I thought it did if we consider $n+ \delta n$ as the closest neighboor of $n$.

I still don't get the point with the subspace. If you consider $\mathbb{R}$ as subspace of itself, why do you use the word subspace then? We can look at it as $\mathbb{R}\subseteq \mathbb{R}^n\; , \;n\geq 1$ and have a subspace topology induced by the one in $\mathbb{R}^n$. So first we need to define the topology of $\mathbb{R}^n$. You said the Euclidean topology, so I assume you mean the standard topology induced by the Euclidean metric. In this case, the subspace topology is also just the ordinary topology of $\mathbb{R}$ which comes from the Euclidean metric. Latest here we can forget about the embedding and subspace topologies. Now I would try to prove that $\mathbb{R}$ is connected.

 

[fresh_42]

Or... to give the sets a better look than those in post #1... $\mathbb{R} = (-\infty,n) \cup [n, \infty)$... intersection is empty...

And $[n,\infty)$ isn't open.

 

[WWGD]

Like Pop'n Fresh pointed out a space with a subspace topology generated by its own open sets just returns the original space's topology.