1. ## Connectedness

Consider the two tangent open discs $\{x^2+y^2<1\}$ and $\{(x-2)^2+y^2<1\}$ in $\mathbb{R}^2$.
How can show that the union of one disc and the closure of the other is a connected subset of $\mathbb{R}^2$?

2. Use the following proposition:

If $A$ and $B$ are connected sets which are not separated, then $A\cup B$ is connected.

Choose $A$ an open disc and $B$ the closure of the other one.

Fernando Revilla

3. Originally Posted by FernandoRevilla
Use the following proposition:

If $A$ and $B$ are connected sets which are not separated, then $A\cup B$ is connected.

Choose $A$ an open disc and $B$ the closure of the other one.

Fernando Revilla
Don't forget to mention that the closure of connected subspaces is connected.

4. Originally Posted by Drexel28
Don't forget to mention that the closure of connected subspaces is connected.
Don't forget that a hint is not a complete solution of a problem.

Fernando Revilla

5. Thanks all. I have another solution using this theorem: If $E$ is a connected subset of $X$and $E\subset A\subset\overline{E}$, then $A$ is connected.

We have $X=E\cup\overline{F}$. Since $E$ is a connected subset of $X$ and $E\subset A=(\{x^2+y^2<1\}\cup(0,-1))\subset\overline{E}$ it follows that $A$ is also connected. Thus $A$ and $\overline{F}$ are connected subsets of $X$such that $A\cap\overline{F}=(0,-1)\neq\emptyset$. Then it follows that $A\cup\overline{F}=E\cup\overline{F}=X$ is connected.

6. That is right, you have used:

Proposition

Let $\mathcal{A}=\{A_i\}$ be a class of connected subsets of $X$ (toplogical space) with non-empty intersection. Then $\big\cup_i A_i$ is connected.

And this proposition is as a consequence that any two members of $\mathcal{A}$ are not separated. All depends on the starting point.

Fernando Revilla