Let $(X,d_{disc})$. Show that $E\subset X$ is disconnected

Question

Let $(X,d_{disc})$ be a metric space with the discrete metric. Let $E$ be a subset of $X$ which contains at least two elements. Show that $E$ is disconnected.

Def: E is disconnected iff $\exists$ disjoint non-empty open sets $V,W \subseteq E$ s.t. $V\cup W=E$

$(\Rightarrow)$ Assume that E is disconnected

WTS $\exists$ disjoint non-empty open sets $V,W \subseteq E$ s.t. $V\cup W=E$

Assume that $V,W \subseteq X$

We are in the discrete metric where each element is $1$ unit apart, therefore $\forall$ $x\in E$ we know that it has to be either in $V$ or $W$ but it can't be in both. This shows that $V$ and $W$ are disjoint.

Also, since there are at least $2$ elements in $E$ we know that we have "enough" elements to ensure that neither $V$ nor $W$ are empty and they are open since the discrete metric is both open and closed, so we can just take its open property.

Intuitively it is clear to me as to why $V\cap W=E$. This is because, say $V$ and $W$ are not "big enough" to equal $E$ we can always make them "bigger" until eventually their union will be equal to E, and they will never overlap due to the above reasoning. But I can't put this into proper words.

Please let me know if this direction is correct so far and any advice on finishing it, so I can complete the other on this post, thank you!

Answer 1

Assume that $V,W \subseteq X$

We are in the discrete metric where each element is $1$ unit apart, therefore $\forall$ $x\in E$ we know that it has to be either in $V$ or $W$ but it can't be in both. This shows that $V$ and $W$ are disjoint.

This is wrong. Why can't $x$ be in both? For all you know, maybe $V=\{x\}$ and $W=\{x,y\}$ for some other point $y$. Or for all you know, maybe $V$ and $W$ are actually the same set! You never assumed they weren't.

The larger issue here is that you are trying to prove the wrong thing. You want to prove there exist $V$ and $W$ with the required conditions (subsets of $X$, disjoint, open, nonempty, their union is $X$). That means you want to describe a specific example of such a $V$ and $W$. You don't want to start by just assuming you have arbitrary $V$ and $W$, since arbitrary $V$ and $W$ probably won't work. For instance, if someone asked you to show that there is a perfect square which is $2$ less than a perfect cube, you wouldn't start by considering an arbitrary perfect square. Most perfect squares aren't $2$ less than a perfect cube! Instead, you would want to just find one specific example that does work: $25$ is a perfect square that is $2$ less than the perfect cube $27$.

So, you want to define some specific subsets $V$ and $W$ of $X$ which have the required properties. As a hint, note that since you want $V\cap W=\emptyset$ and $V\cup W=X$, once you've chosen $V$, you have only one choice of $W$: $W$ must be the complement of $V$ in $X$. So you can think of it instead as picking some subset $V$, such that if you define $W$ to be its complement, then $V$ and $W$ are both open and nonempty. What might you pick $V$ to be?

Answer 2

You are working with the discrete metric, so $\{x\}$ is open, for any $x\in X$.

Take an arbitrary $x_{0}\in E$.

Then, $U=\{x_{0}\}$ is open in $X$, V= $E\setminus \{x_{0}\}$ is open in $X$ (and it is not the empty set, by hyphotesis) and all the properties hold.