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

Question

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.

Answer 1

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.

Answer 2

Option $2.$ is not correct. Consider the set $A=\big\{(x,\sin\big(\frac{\pi}{x}\big):0<x\leq 1\big\}$ 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.