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Is $\mathbb{R}$ closed or open in $\mathbb{R}^2$ with respect to the standard topology of $\mathbb{R}^2$ (open sets are the open epsilon-balls)? I think, $\mathbb{R}$ is closed in $\mathbb{R}^2$ because $\mathbb{R}$ can be identified with $\mathbb{R}\times \{0\}\subset \mathbb{R}^2$ and both sets of the product, $\mathbb{R}$ and $\{0\}$, are closed in $\mathbb{R}$. So, it is a cartesian product of closed sets and therefore closed in $\mathbb{R}^2$ I would say. Is it correct? Regards

toyboy
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3 Answers3

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Formally speaking, $\Bbb R$ is not even a subset of $\Bbb R^2$.

As you say, it can be identified with the subset $\Bbb R\times\{0\}$. But it can also be identified with $\{\pi\}\times\Bbb R$ and with $\Delta=\{(x,x)\mid x\in\Bbb R\}$, and with many many other subsets as well.

Once you choose an identification, you can ask whether or not the set is open or closed. If you chose the $\Bbb R\times\{0\}$, then indeed it is a closed set.

(If by identification you mean finding the image of some embedding as a topological space, then note that $\{0\}\times(0,1)$ can also be identified with $\Bbb R$, which is neither open nor closed.
Note, however, that under this interpretation of "identification" we can at least say that $\Bbb R$ is always $F_\sigma$, since it is $\sigma$-compact, and compact sets must be identified with compact sets again.)

Asaf Karagila
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Let $x_{n}$ in $\mathbf{R} \times \lbrace 0 \rbrace $ be a convergent sequence. Then $x_{n}=(y_{n},0)$ where $y_{n}$ is a convergent sequence in $\mathbf{R}$. Let $y_{n}$ converge to $y \in \mathbf{R}$. $x_{n}$ converges to $(y,0)$, which belongs $\mathbf{R} \times \lbrace 0 \rbrace 0$. Hence the set is closed.

mich95
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As a comment, technically $\mathbb{R}\not\subseteq \mathbb{R}^2$ so we consider $\mathbb{R}\times \{0\}$ as you said.

But yes, you are correct. There are a number of ways to see this. The most direct way to see this is to consider the compliment and draw a picture. Writing down the formal argument shouldn't be hard then. Another way is to consider a convergent sequence in $\mathbb{R}\times \{0\}$ and show it converges to something in $\mathbb{R}\times \{0\}$.