# Let $A$ be a nonempty subset of a topological space $X.$ Which of the statements is true?

Let $$A$$ be a nonempty subset of a topological space $$X.$$ Which of the statements is true?

1. $$A$$ is connected, then its closure $$\overline A$$ is not necessarily connected

2. $$A$$ is path-connected, then its closure $$\overline A$$ is path-connected

3. $$A$$ is connected, then its interior $$A^o$$ is not necessarily connected

4. $$A$$ is path-connected, then its interior $$A^o$$ is connected

My Try: Consider usual topology on $$\mathbb R$$. $$(0,1)$$ is connected. $$[0,1]$$ is also connected. So, 1. is false.

the fourth option is also wrong. Consider the usual topology on $$\mathbb R^2$$.$$\{(x,y)\in \mathbb R^2:(x-1)^2+y^2<1\} \cup \{(x,y)\in \mathbb R^2:(x+1)^2+y^2<1\}\cup \{(0,0)\}$$ is Path-connected. But interior is not connected. So, 3. is correct. Since, It is the opposite of fourth statement.

I am not able to find the counterexample for 2.

• 1 can't be proven wrong with a counterexample. It doesn't say "then the closure is not connected", it says "then the closure is not necessarily connected". Commented Nov 8, 2019 at 6:36
• 3 is not the opposite of 4, however 3 indeed follows from the falseness of 4. Commented Nov 8, 2019 at 7:04

Option $$2.$$ is not correct. Consider the set $$A=\big\{(x,\sin\big(\frac{\pi}{x}\big):0 in $$X=\Bbb R^2$$. Then $$\overline A=A\cup \big(\{0\}\times[-1,1]\big)$$ which is not path connected.
To prove this, note that, $$A$$ is path connected as it is graph of a continuous function on the path connected set $$(0,1]$$.
Next, if possible $$\overline A$$ is path connected. Then there is a path $$\gamma :[0,1]\to A$$ with $$\gamma(0)=(0,0)$$ and $$\gamma(1)=(1,0)$$. So $$\pi_1\gamma,\pi_2\gamma$$ are continuous, $$\pi_1,\pi_2:\Bbb R^2\to \Bbb R$$ are projections on $$1$$-st and $$2$$-nd coordinates. But, $$\pi_1\gamma$$ takes all values of the form $$\frac{1}{n},n\in\Bbb N$$ by intermediate value property of $$\pi_1\gamma$$, as $$\pi_1\gamma(0)=0,\pi_1\gamma(1)=1$$ . So $$\pi_2\gamma$$ assumes each values $$\pm 1$$ in every nbd of $$0\in [0,1]$$. So there is no nbd $$[0,\delta)$$ of $$0$$ in $$[0,1]$$ which maps to $$\big(-\frac{1}{2},\frac{1}{2}\big)$$ under the map $$\pi_2\gamma$$, that is $$\pi_2\gamma$$ is discontinuous.
Regarding 2: Are you familiar with the topologist's sine curve, $$T$$, and its closure?
$$T \smallsetminus \{(0,0)\}$$ is path connected, but its closure, $$\overline{T} = T \cup \{(0,y) : y \in [-1,1]\}$$, is not.