can the empty set contain itself?

Question

This question may sound weird but anyway here it goes. Is the following equation true? $$\emptyset = \{ \emptyset \}$$

The reason I think it is not true:

The empty set is an element by itself, therefore the R.H.S contains one element. But due to L.H.S, the amount of elements of the R.H.S must be $0$.

The reason I think it is true:

The empty set is a subset of every set. Therefore, the emptyset on the R.H.S must be a subset of the L.H.S. In other words, we have the following relation: $\emptyset \supseteq \{ \emptyset \} \Rightarrow \emptyset = \{ \emptyset \}$.

No, it's not true; the empty set is an element of the right hand side, but not of the left hand side. Being an element is not the same as being a subset. — Arturo Magidin, May 20 '19 at 18:49
(Put another way: is an empty bag the same thing as a bag that contains an empty bag inside?) — Arturo Magidin, May 20 '19 at 18:49
I did search after this question before I did poste it. Sorry for dublicate @MauroALLEGRANZA. — LocalMartingale, May 20 '19 at 18:50
You have to separete $\in$ (is an element of) from $\subseteq$ (is a subset of). You have to use "contain" as a synonym of one of the two, but not both. — Mauro ALLEGRANZA, May 20 '19 at 18:51
Indeed. "contain" (or I think "include") is typically used for $\supseteq$, or "owns" for $\owns$. — J.G., May 20 '19 at 18:55

score 0 · Answer 1 · answered May 20 '19 at 18:53

0

The empty set contains itself but the empty set is not an element of itself. The set whose only element is the empty set is not the empty set.

answered May 20 '19 at 18:53

1 Answers1