Given a topological space $(X, \mathcal{T})$, define $\trianglelefteq_{\mathcal{T}} = \{(a,b) \in X^2 : a \in \overline{\{b\}}\}$.
Prove that $\trianglelefteq_{\mathcal{T}}$ is a preorder, that is, transitive and reflexive.
Here is my understanding of the problem:
We need to show that it is both reflexive and transitive.
- $a \trianglelefteq_{\mathcal{T}} a$ for all $a$ since $(a,a) \in X^2$
- $a \trianglelefteq_{\mathcal{T}} b$ and $b\trianglelefteq_{\mathcal{T}} c$ implies that $a \trianglelefteq_{\mathcal{T}} c$.
I am unsure about this one, but here is what I've got:
if $a \trianglelefteq_{\mathcal{T}} b$, then we have that $(a,b) \in X^2$ and if we are also given that $b \trianglelefteq_{\mathcal{T}} c$, then $(b,c) \in X^2$. However, we are told that $a \in \overline{\{b\}}$, and by the second relation between $b$ and $c$, we would have $b \in \overline{\{c\}}$.
At this point, I want to say that $b = \overline{\{b\}}$ and use that to associate $a$ and $c$; however, may we assume that $b = \overline{\{b\}}$, or would we need a new direction to proceed in this manner?
Much thanks in advance for your attention to this question.
To show that $a\trianglelefteq_{\mathcal{T}}a$, it’s not enough to say that $\langle a,a\rangle\in X^2$: you must also point out that $a\in\operatorname{cl}\{a\}$.
You certainly can’t say that $b=\operatorname{cl}\{b\}$, for two reasons. First, $b$ is a point of $X$ and $\operatorname{cl}\{b\}$ is a subset of $X$, so they aren’t even the same kind of object: at best you could have $\{b\}=\operatorname{cl}\{b\}$. However, there is no reason to suppose that this is actually the case. It’s entirely possible, for instance, that $\operatorname{cl}\{b\}=X$.
For transitivity you’re assuming that $a\trianglelefteq_{\mathcal{T}}b$ and $b\trianglelefteq_{\mathcal{T}}c$, so by definition you know that $a\in\operatorname{cl}\{b\}$ and $b\in\operatorname{cl}\{c\}$. You want to show that $a\trianglelefteq_{\mathcal{T}}c$, i.e., that $a\in\operatorname{cl}\{c\}$. It suffices to show that $\operatorname{cl}\{b\}\subseteq\operatorname{cl}\{c\}$; can you see why this is true?