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I want to prove $C=\{v=(x,y)|x,y\geq 0\}$is a closed subspace of $\mathbb R^2$.

Basically I want to generalize the suggestion in this post for any $\mathbb R^n$:

Do I need to show that $v_n \underset{n \to \infty} \to (x,y)=v\in C$ for all $v_מ$?

What will be a good system for this kind of proofs?

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    $\begingroup$ Subset would be a better word here than subspace $\endgroup$
    – zhw.
    May 22, 2016 at 19:58

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What you have shown is that for any $v\in \mathcal{C}$, there is a sequence $(v_n)$ of elements of $\mathcal{C}$ which converges to $V$. For $\mathcal{C}$ to be closed, it should be that any convergent sequence of elements of $\mathcal{C}$ converges in $\mathcal C$. I suggest another way: notice let $f:\mathbb{R}^2\mapsto \mathbb{R}^2, (x,y) \mapsto (x,y)$ be the identity. It is obvious $f$ is continuous, furthermore notice $\mathcal{C}=f^{-1}[0,+\infty)\times[0,+\infty)$. But $[0,+\infty)\times[0,+\infty)$ is closed for the product topology, standard on $\mathbb{R^2}$, hence $\mathcal C$ is closed.
In fact this is a long winded tautology, since $\mathcal C$ is already defined as $[0,+\infty)\times[0,+\infty)$: just notice it is the cartesian product of two closed sets, hence closed for the product topology (the fact that $[0,+\infty)$ is closed is what the post you refer to asked to prove).

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  • $\begingroup$ Why is $C$ defined as $[0,+\infty )\times [0,+\infty )$? I am not familiar with this notation. all i know is $c=\{(x,y)|x,y\geq 0\}$. how does it translate to $\times$? $\endgroup$
    – Maxim
    May 24, 2016 at 13:24
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    $\begingroup$ @Maxim $(a,b)\times (c,d)$ means you are taking the direct product of the two intervals that form ordered pairs of the form $\{(x,y): a< x <b, c<y<d \}$ Also, $\mathbb{R}^2$ is an abbreviation for $\mathbb{R}\times \mathbb{R}$ $\endgroup$ May 24, 2016 at 16:21
  • $\begingroup$ I don't understand then, why is it suitable here? if I understand correctly $C$ is containing ordered pairs but it's element are not a direct product of ordered pairs. What am I missing? $\endgroup$
    – Maxim
    May 24, 2016 at 16:25
  • $\begingroup$ can this be applied to $\mathbb R^3$? Can I say that a $C=\{(x,y,z)|x,y,z,\geq 0\}$ is also closed since it is defined as $[0,+\infty )\times [0,+\infty )\times [0,+\infty )$? $\endgroup$
    – Maxim
    May 24, 2016 at 16:30
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    $\begingroup$ In general for sets $A_1,A_2,..,A_n$, we can define the Cartesian product $A_1\times A_2\times...\times A_n=\{(x_1,x_2,...,x_n), x_i \in A_i, 1\leq i\leq n\}$, the set of ordered $n$-tuples with one component for each of the $A_i$'s. (this is also where the notation $\mathbb{R}^2=\mathbb{R}\times \mathbb{R}$ comes from). As you noticed, $\mathcal{C}=\{(x,y), x,y \geq 0\}=\{(x,y), x,y\in [0,+\infty)\}$. You can indeed extend this to any of the $\mathbb{R}^n$. $\endgroup$ May 24, 2016 at 18:21

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