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This question is inspired by thermodynamics. To give the setting, we first must understand the basic principle behind the second law.

Suppose you have two systems: system $A$ at temperature $T_1$ and system $B$ at temperature $T_2$. Denote the flow of internal energy from system $A$ to system $B$ during a (short) moment Δt by ΔU (so that ΔU is positive if it flows from $A$ to $B$). We therefore say, by convention, that positive ΔU means energy flows A → B, negative ΔU means B → A.

The guiding equation for the behavior of the combined system will be

$$ \Delta S = - \Delta U/T_1 + \Delta U/T_2 \geq 0.$$

The above law is just what physicists refer to the second law of thermodynamics. *

We want to define a “thermodynamic-temperature line” from the fact that

$$ \Delta S\ge0\;\Longrightarrow\;\Delta U \left(\frac 1{T_2} -\frac1{T_1}\right) \ge 0, $$

or

$$ \operatorname{sign}(\Delta U)=\operatorname{sign}\left(\frac1{T_2} -\frac1{T_1}\right). $$

An interesting question is what this translates to in topology when we try to "order" temperatures by how energy would flow between system $A$ and system $B$. Consider now an extended real‐line

$$ \widetilde{\mathbb R} := [-\infty, \, +\infty] \cup \left \{ 0_-, \, 0_+ \right\} $$

with the strict order

$$ T_1\succ T_2\quad\Longleftrightarrow\quad \Delta U>0 \quad \Longleftrightarrow \frac 1{T_2} > \frac 1{T_1}. $$

To make the thermodynamic "temperature scale" work, it could better to have two zeros: $0_-$ and $0_+$ with the usual ordering $0_- < 0_+$. We also add the rules that

$$\begin{cases}1/0_- = -\infty, \\ 1/0_+ = +\infty, \\ 1/(- \infty) = 0_-, \\ 1/(+ \infty) = 0_+. \end{cases}$$

Since

$$-\infty = 1/0_- < 1/0_+ = +\infty$$

we also have that $0_- \succ 0_+$.

Having two zeros might be weird, but we only care insofar about the ordering on $\widetilde {\mathbb R}$.

Suppose we induce the order topology on $\widetilde {\mathbb R}$ with this given order $\succ$, i.e. the topology will be generated from the basis consisting of "intervals"

$$\left\{ x : a < x < b \right\}.$$

We can call this – what I think is appropriate to call it – the “thermodynamic order‐topology”, named after its origin in statistical mechanics. Is there a way to characterize open sets in this topology?

Will this topological space be Hausdorff? Compact? Connected? Which maps $f \colon \widetilde {\mathbb R} \to \widetilde {\mathbb R}$ are continuous?

* a caveat is that this analysis assumes the systems do not exchange anything but internal energy and are kept at their pressures and volumes, so no work is done by external forces (and of course the two systems have to be thermodynamically isolated from the rest of the universe).

Markus Klyver
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  • Really interesting post, I never think about thermodynamic like this haha. Only a point $\frac{1}{T_2} > \frac{1}{T_1}$ iff $T_1 > T_2$ so your order is like the usual order: $T_1 \succ T_2$ iff $T_1 > T_2$ where $>$ is the usual order. For this I must be possible to answer your questions. – Brian Britos Simmari Jun 10 '25 at 22:51
  • What is this order supposed to be? – Jakobian Jun 10 '25 at 23:23
  • How will you define comparisons with $0$? $1/0$ can be thought of as an element of the one point compactification of $\mathbb R$, but not as an element of the two point compactification. – Cheerful Parsnip Jun 11 '25 at 02:23
  • @CheerfulParsnip You are correct I need to be more careful here. We can solve this problem by introducing two zeros: $0_-$ and $0_+$. I expand on this in my updated/edited question. – Markus Klyver Jun 11 '25 at 12:02
  • This is still incoherent – Jakobian Jun 11 '25 at 19:27
  • @Jakobian The order $\succ$ is defined in terms of the usual one. If you don't like the symbols $T_1$ and $T_2$, we can replace them with $x$ and $y$ instead. Suppose we have the usual order of $\mathbb R$, with the additions of $0_-$ and $0_+$ (so that we get $\widetilde {\mathbb R}$. We then define a new order $\succ$ by $x \succ y$ iff $1/y > 1/x$. We can therefore equip this ordered set with the order topology, and the topological questions follow. Is there a particular piece of information you still think is unclear, or needs clarification? – Markus Klyver Jun 11 '25 at 20:24
  • If I understand it correctly then, you are defining $\tilde{\mathbb{R}} = ([-\infty, \infty]\setminus {0})\cup {0^+, 0^-}$ with the order $[-\infty, 0^-]$ the same as $[-\infty, 0]$, the order on $[0^+, \infty]$ the same as $[0, \infty]$, and $0^- < 0^+$? If that's the case, this space $\tilde{\mathbb{R}}$ is not interesting at all, really, its just a union of closed intervals. – Jakobian Jun 11 '25 at 22:55
  • @Jakobian What's interesting is not $\left( \widetilde {\mathbb R}, , > \right)$ but rather $\left( \widetilde {\mathbb R}, , \succ \right)$. – Markus Klyver Jun 12 '25 at 08:19
  • @MarkusKlyver By what you explained its the same thing – Jakobian Jun 12 '25 at 08:42
  • @Jakobian Is it really that obvious that $\left( \widetilde {\mathbb R}, , > \right) \cong \left( \widetilde {\mathbb R}, , \succ \right)$ as ordered fields? – Markus Klyver Jun 12 '25 at 21:11
  • @MarkusKlyver maybe once we agree on what $\succ$ is – Jakobian Jun 12 '25 at 21:19
  • @Jakobian I define $\succ$ in terms of $>$. I say that $x \succ y$ if $1/y > 1/x$. Please tell me if this is unclear. – Markus Klyver Jun 12 '25 at 21:57
  • @MarkusKlyver and what is $1/\pm \infty$? $0_{\pm}$? – Jakobian Jun 13 '25 at 11:09
  • @Jakobian That is correct. – Markus Klyver Jun 16 '25 at 20:02

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