Is the closure of $S^1-\{(0,1)\}$ compact?

Question

My question is very simple : Is the closure of $S^1-\{(0,1)\}$ compact ? (where $S^1\subset \mathbb R^2$ is the unit circle). In somehow, I think it is, since it should be the circle, but in on other way, $S^1-\{(0,1)\}$ is homeomorphic to $\mathbb R$, so in somehow, if the closure of $S^1-\{(0,1)\}$ is compact, then so is the closure of $\mathbb R$ (which I guess it's not). So, at the end, I'm not so sure if the closure of $S^1-\{(0,1)\}$ is really compact. Any idea ?

Answer 1

Remember the notion of closure depends on the surrounding space (i.e., the phrase "closure of $X$" is not a topological notion: it depends on more than just the topology of $X$, but also on how $X$ is embedded in some larger space).

So we can't talk about the closure of $S^1 \setminus \{(0, 1)\}$ or of $\Bbb{R}$ without knowing what surrounding space is intended. Here, in the original question, it is implicit that $S^1 \setminus \{(0, 1)\}$ is being thought of as a subspace of $\Bbb{R}^2$ in the standard way, which means the closure is the compact space $S^1$. When we talk about $\Bbb{R}$ we are usually thinking of it as a space in its own right (or perhaps as a subspace of $\Bbb{C}$) and, on that understanding, it is its own closure, so the closure is not compact.

Answer 2

You are expecting too much about the closure operation. Take $S^1-\{(0,1)\}$, $\mathbb{R}$ e $(0,1)$. The closure of the first one is $S^1$, of the second one is $\mathbb{R}$ and the last is $[0,1]$. The closure are not homeomorphic.

There is a reason for that: the closure is a extrinsic property. Look at the definition: we start with a topological space $X$ and a subset $S$. The closure of $S$ at $X$ is the intersection of all closed sets of $X$ that contains $S$. It's clear from the definition that the closure of $S$ depends of the whole topological space, not the set on is own.

So, I will rewrite what I did write better:

• The closure of $S^1-\{(0,1)\}$, as a subset of $\mathbb{R}^2$, is $S^1$
• The closure of $(0,1)$, as a subset of $\mathbb{R}$, is $[0,1]$
• The closure of $\mathbb{R}$, as a subset of $\mathbb{R}$, is $\mathbb{R}$

See? We are working with non homeomorphic topological spaces, that's why you can't say the closure is preserved by homeomorphism.

To make it more clear, look what happen when we consider $S^1-\{(0,1)\}$ as a subset of $S^1-\{(0,1)\}$: in this case, $S^1-\{(0,1)\}$ is already closed, and then its closure is $S^1-\{(0,1)\}$. The closure changes when we change the topological space.

So, answering your initial question:

• The closure of $S^1-\{(0,1)\}$, as a subset of $\mathbb{R}^2$, is compact.
• The closure of $S^1-\{(0,1)\}$, as a subset of $S^1-\{(0,1)\}$, isn't compact.

Answer 3

The closure of $S^1 - \{(0,1)\}$ as a subspace of $\mathbb{R}^2$ is the circle. Since it is closed and bounded we know it is compact. As other users pointed out, you need to consider the topological space involved, because the notion of closure depends on the space, but also on the topology of the space.

Regarding homeomorphisms, the circle is homeomorphic to the one point compactification of the real line $\mathbb{R}$. As you said, $S^1 - \{(0,1)\}$ is homeomorphic with the real line and both spaces are Hausdorff and locally compact. For this reason you can extend this homeomorphism to another homeomorphism between the one point compactifications, that is between $S^1$ and $\mathbb{R} \cup \{\infty\}$. Hence, we conclude again that $S^1 = Cl(S^1 - \{(0,1)\})$ is compact.