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How would I go about showing that $ \mathbb{R} $ with the topology $ \tau = \{ \emptyset \} \cup \{ U | \pi \in U \subseteq \mathbb{R}\} $ is connected?

Preferably an explanation someone who isn't good at topology could understand. Not sure where to start with this. I understand the being connected is a topological property and that something is connected if it;s not disconnected.

A topological space is disconnected if if there exist $ U, V \in \tau $ which are nonempty and disjoint with X = U ∪ V

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  • $\begingroup$ Can you find two nonempty disjoint sets... at all? $\endgroup$ – G Tony Jacobs Dec 11 '17 at 15:15
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So the idea is that you want to show that if you have two open sets $U$ and $V$ such that $U \cap V = \emptyset$ and $U \cup V = \mathbb{R}$, then one of either $U$ or $V$ must be the empty set.

Here's a big hint: in order for $U \cup V$ to be equal to $\mathbb{R}$, at least one of $U$ or $V$ must contain the element $\pi$ (suppose wlog that it's $U$). As $U$ and $V$ are open sets, hence in the set $\tau$, what does the property that $U \cap V = \emptyset$ mean the set $V$ must be?

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  • $\begingroup$ Then V must be the empty set which forms a contradiction sicne U and V are non-empty by hypothesis - assuming it's disconnected? Since it can't be disconnected -> conencted. $\endgroup$ – Overflow2341313 Dec 11 '17 at 15:21
  • $\begingroup$ Technically the precise way I've set up the definition of connected means you can stop at the 'hence $V$ is empty' step, but yes that's correct. $\endgroup$ – Dan Rust Dec 11 '17 at 15:23
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Hint: If the set $\mathbb{R}$ would not be connected in this topology, the definition tells you that then you could find two nonempty, open and disjoint sets $U_1$ and $U_2$ such that $\mathbb{R} = U_1 \cup U_2$.

By the definition of what open means in your case (recall that open = set is in the topology), you should be able to get a contradiction from this. Then you have proven by contradiction that the set is connected.

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