# Freddy Krueger's Lullaby

1, 2.

Freddy is drawing eyeballs.

2½, 3.

One iris got bigger.

3½, 4.

The other got bigger.

5, 6.

Aura passes from and to irises.

[NUMBERS REDACTED]

Never sleep again.

Subtle Hint:

Moderate Hint:

The numbers are redacted because currently, numbers end by 6.

Decisive Hint:

The "aura" may seem like equipotentials between heterogeneously charged conductors: 1, 2. Freddy is drawing eyeballs.

In $$T_2$$ spaces, or Hausdorff spaces, every two distinct points can be seperated by open sets. It is commonly depicted by the following diagram: 2½, 3. One iris got bigger.

In $$T_3$$ spaces, or regular spaces, every disjoint closed set and point can be seperated by open sets. It is commonly depicted by the following diagram: 3½, 4. The other got bigger.

In $$T_4$$ spaces, or normal spaces, every two disjoint closed sets can be seperated by open sets. It is commonly depicted by the following diagram: 5, 6. Aura passes from and to irises.

In $$T_6$$ spaces, or perfectly normal spaces, for every two disjoint closed sets, there exists a continuous function whose return value is 0 on and only on one closed set and 1 on and only on the other closed set. (Hence the decisive hint)

• Great! This probably should be the accepted answer (regarding the existence of $T_{2\frac12}$ and $T_{3\frac12}$). – trolley813 Jan 6 at 22:08
• @trolley813 Well, it is the OP self-answering - I'd very much hope this was correct! :) – Stiv Jan 6 at 22:25
• For those of us too slow to make the connection ourselves, can you add an explanation of how Freddy Krueger fits in? – tmpearce Jan 9 at 16:34

Cassini ovals? (or some similar curves)

Reasoning

The equation of the oval is $$(x^2+y^2)^2-2a^2(x^2-y^2)+a^4=b^4$$. When the $$c = \dfrac b a$$ ratio increases, the oval initially looks like "eyeballs" (2 separate loops, indeed starting with 2 dots when $$c=0$$), with increasing "irides", then (when $$c=1$$) the halves are fusing with each other (it will be Bernoulli lemniscate - an 8-shaped curve), allowing the "aura" to pass between them, and next (when $$c>1$$) turning into single peanut-shaped (later, when $$c>\sqrt2$$, an oval-shaped) curve.

However

this answer bears absolutely no connection with A Nightmare on Elm Street franchise, and it probably has to have one (since Freddy Krueger is the protagonist name, and Never Sleep Again is a documentary film that chronicles the entire franchise, and the number of installments in the main series is 6 at the moment (currently numbers end by 6!)).

• Ha, nice and creative guess, but not what I intended. I intended some other branch of math. You should consider why there are numbers like 2 1/2. – Dannyu NDos Dec 30 '19 at 7:12
• Hint: I didn't intend much connection to the franchise. – Dannyu NDos Dec 30 '19 at 7:13
• As your answer was the only one, I gave you the bounty. – Dannyu NDos Jan 5 at 20:17
• @DannyuNDos Thanks! It's a pity that nobody gave a better answer. – trolley813 Jan 5 at 21:36