Can something contain iteself? [closed]

I asked this over on the Phyisics part of StackExchange, and they suggested I move my question here.

And said question is:

Can something contain itself?

The question is simple enough, and I can think of two ways it could happen:

1: The object contains a portal to itself

This is a kind of a loophole, but still may be impossible, so I'm including it.

Let's start with Box A. Inside of Box A, we have Box B, which is a portal to Box A: Portals on the outer sides of Box B correspond to the respective inner sides of Box A. Note that I said portals, when really the idea is of one, continuous portal.

For extra interesting-ness, What if the sides on Box B didn't correspond to Box A's sides? What if it was random? Could the portal still be continuous or would it have to be 6 differing portals, one for each side?

Are either of the above ideas Mathematically or Physically possible?

2: The object literally contains itself

Box A doesn't contain Box B anymore. Now, instead of them being two different boxes, they're both the same. Let's calle them Box A$_0$ instead of Box A and, inside of Box A$_0$, there is Box A$_1$. This raises a few questions:

• Is this possible?
• Would this go on forever, where Box A$_1$ contains a Box A$_2$, which contains a Box A$_3$
• If this is so, would Box A$_0$ be inside of a Box A$_-1$, which would, again, go on forever?

Note:

I tagged this as topology as I wasn't totally sure as to what it should belong in. Suggestions?

closed as off-topic by Henning Makholm, Mnifldz, Andreas Blass, Asaf Karagila♦, Austin MohrMay 26 '15 at 20:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – Henning Makholm, Mnifldz, Andreas Blass, Asaf Karagila, Austin Mohr
If this question can be reworded to fit the rules in the help center, please edit the question.

• I don't know if this is a "math" question or not. In math terms, the "set containment" is reflexive, meaning for any set $A$, $A\subseteq A$ (meaning that $A$ is contained in $A$). – TravisJ May 26 '15 at 19:48
• I don't know what they told you at Physics, but this is certainly not a mathematical question either. It seems to be a philosophical one, or perhaps one of what exactly you mean when you say "contain". – Henning Makholm May 26 '15 at 19:48
• In set theory the axiom of foundation guarantees that this cannot happen. – Mnifldz May 26 '15 at 19:50
• That is, a set cannot contain itself as a member. It can (and does) contain itself as a subset. – Robert Israel May 26 '15 at 20:04
• If you want to ask a mathematical question, it helps to know precisely what you mean by the question. – Robert Israel May 26 '15 at 20:19

Yes, in mathematics it is quite easy to take spaces and cut or glue as you desire. (Manifesting such objects physically might be difficult or impossible, but that's not mathematics.)

Here's how I interpreted both of the situations you describe (I'd say they're really the same). Represent “box $A$” as the space $I^3$, where $I=[0,1]$, and “box $B$” as the subspace $J^3\subset I^3$ where let's say $J=[\frac{1}{4},\frac{3}{4}]$. Then you're just

• removing the interior of box $B$

• identifying the faces of box $B$ with box $A$

which produces a perfectly reasonable space, homeomorphic to $S^2\times S^1$. It's easier to see this if you look at the two-dimensional analog first:

I'd emphasize that such a construction is quite mundane in mathematics, it doesn't warrant such serious talk of whether it is “mathematically possible”.

I'd also emphasize that, after interpreting your question in this way, it becomes clear that it makes no sense to discuss whether “$A$ really contains $B$” (or whether “$A$ really contains $A$”), nor whether there's anything about this situation which “goes on forever”. It is important to get rid of all terminological / philosophical confusion.

• About the last paragraph, I meant as in if A$_0$ contained A$_1$, then, since they're the same box, it must follow that A$_0$ is also inside of something. – Quelklef May 26 '15 at 21:11
• There is no "inside" / "outside" / "contains" / "contained in" once you've made the identification of the faces. – Zev Chonoles May 26 '15 at 21:21