Translate “any open interval” and “any closed interval” from English to math symbols.

first-order-logiclogicpredicate-logic

In my intro to real analysis book, I came up with the following lemma (which is easy to prove) to help with an exercise.

For any open interval $K$, if any closed interval $S$ that is a subset of $K$ has the property that for any $x \in S$:$\varphi(x)$, then for any $x\in K$, we must have $\varphi(x) \quad(\dagger)$

For the proof, consider any $x \in K$. Now, consider the set $\{x\}$. This set is a closed set. Therefore, by assumption $\varphi(x)$.

Question: Could someone help me translate the English of $(\dagger)$ into the appropriate math notation?

How does one encode "$K$ is any open interval" and "$S$ is any closed interval"? Are these strictly topological notions that require new notation? Or is there a clever way that uses basic quantifiers and inequalities?

For example, does the following syntax work?

$\forall K \Bigg[\bigg(\Big[\exists a,b \in \mathbb R: \forall x (a \lt x \lt b \rightarrow x \in K) \Big] \text { and } \Big[\forall S\color{blue}{\big(}\color{red}{(}\exists a,b \in K: \forall x (a \leq x \leq b \rightarrow x \in S)\color{red}{)}\rightarrow \forall x \in S: \varphi (x)\color{blue}{\big)}\Big]\bigg) \rightarrow \forall x \in K: \varphi (x) \Bigg] $

Edit:

I think it may be necessary to add a further specification to each of the conjuncts in the overarching antecedent.

For example, in the statement $\exists a,b \in \mathbb R: \forall x (a \lt x \lt b \rightarrow x \in K)$, I have not ensured that $K=\{x \in \mathbb R : a \lt x \lt b\}$. Rather, I have only ensured that $\{x \in \mathbb R : a \lt x \lt b\} \subseteq K$. To guarantee equality, I would have to add the condition that $\forall x ( x \leq a \text{ or } x \geq b \rightarrow x \notin K)$.

A similar extra condition would have to be stipulated for $S$.

Best Answer

I want to make a few notes:

  • Bear in mind that stuff like $K, S$ are just names. You don't need to give coordinates/bounds for your intervals necessarily, unless you want to specifically reference those bounds. It's the same with any set: unless you need to reference particular features or elements, don't overcomplicate details.

  • You can assume, as needed, any definitions in statements like these. Your translation need not somehow include a definition of what it means to be an open interval. If need be, you can literally just say what $K,S,$ etc. are, if doing so is cumbersome symbolically. There generally isn't issue with this. Mathematical language is in part about communication, after all.

    • (After all, going to an extreme, we shouldn't have to define basic notions like set inclusion, set membership, or even basic notations like "what is a real number" for stuff like this to be meaningful. Those are handled elsewhere; make sufficient assumptions on what the reader knows so that we can concisely yet precisely communicate what we need to.)

So, to translate

For any open interval $K$, if any closed interval $S$ that is a subset of $K$ has the property that for any $x \in S$:$\varphi(x)$, then for any $x\in K$, we must have $\varphi(x) \quad(\dagger)$

I would simply go with

$$\begin{align*} &(\forall K \subseteq \Bbb R \text{ an open interval})(\forall S \subseteq K \text{ a closed interval})(\forall x \in S)(\varphi(x)) \\ &\implies (\forall x \in K)(\varphi(x)) \end{align*}$$

Broken down a bit:

  • Let $K \subseteq \Bbb R$ be some open interval...
  • ...where for any subset $S \subseteq K$ where $S$ is a closed interval...
  • ...and each $x \in S$ has property $\varphi(x)$.
  • Then we say that property $\varphi(x)$ holds for all $x \in K$.

If you want to avoid the writing things as text, you could then introduce notation representing $K,S$ as intervals formally: $K = (k,\ell)$ and $S = [s,t]$ below:

$$(\forall (k,\ell) \subseteq \Bbb R)(\forall [s,t] \subseteq (k,\ell) )(\forall x \in [s,t])(\varphi(x)) \implies (\forall x \in (k,\ell))(\varphi(x))$$

(Or something to this effect anyhow, it's hard for me to parse through what you wish to translate, but I think I'm somewhere close at least.)

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