Questions on connectedness

Question

I have a final examination in general topology this week, and I've been doing past papers for the past two days in anticipation for it. I'm not sure if my answers are correct so could someone tell me if I'm in the right direction?

Let $(X,\tau_1)$ and $(Y, \tau_2)$ be topological spaces

$1) \quad\text{Prove that $X$ and $Y$ connected iff $X\times Y$ connected}$

$\text{Sol$^n$: X and Y connected $\longleftrightarrow$ $\{x\in X|\quad\text{X is connected}\}$}\text{and {$y\in Y$| $\text{Y is connected}$}}$

$\longleftrightarrow \{(x,y)\in X\times Y |\quad \text{X is connected and Y is connected}\}$

$\longleftrightarrow X\times Y\quad \text{is connected}$

$2)$ Prove that X is connected iff for each pair of open subsets in $U$ and $V$ of $X$, we have $U\subseteq V$ or $V\subseteq U$

$\text{Sol}^n: $ $\longrightarrow$ Since $X$ is connected we know for $U,V\in X,\quad$ $U\cup V = X$ and $U\cap V \neq\emptyset$. Since $U\cap V \neq\emptyset$ the implication here is that $U\subseteq V$ or $V\subseteq U$

$\longleftarrow$ If for $U,V\in X$ we have $U\subseteq V$ or $V\subseteq U$ then it is clear that $U\cup V = X$, and $U\cap V \neq \emptyset$ so X will be indeed connected

$3)$ Let $(\mathbb{R},\epsilon(0))$ be a topological space with topology

$$\epsilon(0) = \{\mathbb{R}\}\cup\{S\subseteq\mathbb{R}: 0\not\in S\}$$

Prove that $(\mathbb{R},\epsilon(0))$ is connected.

I'm having a bit of trouble with this one. If we exclude 0 as in the definition then wouldn't this be a partition of the set such that for $U,V\in \epsilon(0)$, $U\cap V=\emptyset$?

Answer 1

Your answer for (1) isn't adequate. The second equivalence is just a restatement of the result you are trying to prove. Also, the result doesn't hold unless both $X$ and $Y$ are assumed to be nonempty.

To prove this equivalence, one direction is easy: If either $X$ or $Y$ is disconnected, then it is trivial to show that $X\times Y$ is also disconnected.

For the reverse direction, here's one possible argument:

Suppose that $X\times Y=U\cup V$ with $U$ and $V$ nonempty and disjoint. We show that we cannot have $U$ and $V$ both open. Since $U$ and $V$ are nonempty, we can, without loss of generality, find $x_0,x_1,y_0$ such that $\langle x_0,y_0\rangle\in U$ and $\langle x_1,y_0\rangle\in V$. (The "WLOG" is necessary since we may need to swap the roles of $X$ and $Y$).

Intuitively, we're going to consider the "line segment" connecting $\langle x_0,y_0\rangle$ and $\langle x_1,y_0\rangle$. This must cross $U$ to $V$, which will be difficult if $U$ and $V$ are both open. It isn';t a real line segment, but it must be "parallel" to the $x$-axis.

Consider the sets $A\subseteq X$ given by $$\big\{a\in A \big\vert \langle a,y_{0}\rangle\in U\big\}$$ and $B=X\setminus A$. Clearly $x_0\in A$ and $x_1\in B$. Now, since $X$ is connected, we can't have $A$ and $B$ both open, which means that there is a point $a\in A$ such that $a\in\partial A$ (the boundary of $A$). Now consider the point $\langle a,y_0\rangle$. If you can show that it is on the boundary of $U$ (which means simply that every neighborhood of $\langle a,y_0\rangle$ intersects both $U$ and $V$), then you will have shown that the boundary of $U$ is nonempty, and thus that $U$ cannot be both open and closed.

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Problem (2) is misstated. It is clearly not true.

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For the space in (3) to be disconnected, you'd need two nonempty disjoint open sets $U$ and $V$ whose union is the whole of $\mathbb{R}$. But the only open set that contains $0$ is the whole space $\mathbb{R}$. Therefore of one $U$ and $V$ must be $\mathbb{R}$. But in this case, how can the other be nonempty?