Showing set is closed and other sets not closed. For $i=1$ and $i=2$, consider the mappings $q_i$:  $\mathbb{R^2} \to \mathbb{R}$ defined by $q_i$($x_1$,$x_2$)=$x_i$ ( that is, $q_1$ and $q_2$ are projections on the x and y axis respectively).  Please show that the set G:={($x_1$,$x_2$) $\in$ $\mathbb{R^2}$, $x_1$$x_2$=1} is closed in $\mathbb{R^2}$ but neither $q_1$(G) nor $q_2$(G) are closed in $\mathbb{R}$.
Thoughts/Attempt to think about it: I know how to show a set is generally closed(have done so many times) either proving that the set contains its limit points or that the complement of the set is open.  How do I show that the complement of the set G is open?  I don't really have much intuition for this problem as I haven't encountered or seen a problem like it in any text (this is my first introduction to analysis/topology).  Generally, to show a set is open I show that all the points are interior points.  That is, for each point in the set G complement, I show that there exists a ball around it that is fully contained in the set.  However, I don't know how to formalize this argument to this particular question or if this is even the right approach (should I consider limit points instead?)
Also I know that if  neither $q_1$(G) nor $q_2$(G) are closed in $\mathbb{R}$, that does not mean that they are open, they could be clopen.   If I show that they are indeed open, then neither are closed.  However I do not think that this is necessarily the best strategy as I don't know beforehand whether they are clopen or open.  
Any help, guidance, instruction, pointers would be much welcomed.
Thanks!
 A: I pressume that you use te Euclideian metric $d_2$ in $\mathbb{R}^2$ where $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$
Let $x_n=(x_n^1,x_n^2) \in G$ such that $x_n \to^{d_2} x=(x_1,x_2)$
Then it is not difficult to see that $$x_n^1 \to x_1$$ $$x_n^2 \to x_2$$
Thus $1=x_n^1x_n^2 \to x_1x_2$,so from uniqueness of limit we have that $x_1x_2=1$ so $x=(x_1,x_2) \in G.$
Thus from the sequentail  characterization of closedness we prove that $G$ is closed.
Now if $(x_1,x_2) \in G$ we have that $x_1,x_2 \neq 0$ and $x_1=\frac{1}{x_2}$
So $q_1(G)=\mathbb{R}^2 \setminus \{0\}$ which is an open set as a complement of a singleton which is a closed set in $\mathbb{R}^2$
Apply the same argument for $q_2(G)$
A: Hint:
$$q_1(G) = q_2(G) = \mathbb{R} \setminus \{0\} $$
A: (I). A set can be open or closed or both or neither. The definition of a closed set is the complement of an open set.
(II).In $\Bbb R$ the definition of the usual (standard) topology is that  $S$ is open iff for each $x\in S$ there is an $r_x>0$ such that $(-r_x+x,r_x+x)\subset S.$ In particular $\emptyset$ is open (as there is no $x$ in $\emptyset$ that violates the condition for being open) and $\emptyset$ is closed because its complement $\Bbb R$ is open.
In $\Bbb R$ there are other, equivalent ways to define closed sets.
(III). The projection  of $\{(x,y)\in \Bbb r^2:xy=1\}$ onto either its $1$st or $2$nd co-ordinate is $T= \Bbb R \setminus \{0\}.$ Now $T$ is open because if $0\ne x\in \Bbb R$ then $|x|/2>0$ and $0\ne (-|x|/2+x, |x|/2+x)$ so $(-|x|/2+x,|x|/2+x)\subset T.$ And also $\Bbb R \setminus T=\{0\}$ is closed.
(IV). In $\Bbb R^2$ there are several (equivalent ) ways to define the usual (standard) topology. For example: 
(a).For $p=(x,y)$ and $p'= (x',y')$ in $\Bbb R^2$ define $d(p,p')=\sqrt {(x-x')^2+(y-y')^2}\;.$ 
For $0<r\in \Bbb R$ and $p\in \Bbb R^2$ the set $B_d(p,r)=\{p'\in \Bbb R^2: d(p,p')<r\}$ is a circular region with center $p$ and radius $r.$  
Define $S\subset \Bbb R^2$ to be $d$-open iff for each $p\in S$ there is an $r_p>0$ such that $B_d(p,r_p)\subset S.$ 
(b). For $p=(x,y)\in \Bbb R^2$ and $p'=(x',y')$ let $e(p,p')=\max (|x-x'|,|y-y'|).$ For $0<r\in \Bbb R$ and $p\in  \Bbb R^2$ the set $B_e(p,r)=\{p'\in \Bbb R^2: e(p,p')<r\}$ is a square region with sides parallel to the co-ordinate axes, with center $p,$ and with side-length $2r.$ 
Define $S\subset \Bbb R^2$ to be $e$-open iff for each $p\in S$ there is an $s_p>0$ such that $B_e(p,s_p)\subset S.$ 
(c). $d$-open and $e$-open are equivalent:
(c-1). If $S\subset \Bbb R^2$ is $d$-open is open and $p\in S$  take $r_p>0$ such that $B_d(p,r_p)\subset S.$ Let $s_p=r_p/2.$ Then $s_p>0$ and $B_e(p,s_p)\subset B_d(p,r_p)\subset S.$ Therefore if $S$ is $d$-open then $S$ is $e$-open. 
(c-2). If $S\subset \Bbb R^2$ is $e$-open  and $p\in S$  take $s_p>0$ such that $B_e(p,s_p)\subset S.$ Let $r_p=s_p.$ Then $r_p>0$ and $B_d(p,r_p)\subset B_e(p,s_p)\subset S.$ Therefore if $S$ is $e$-open then $S$ is $d$-open. 
Being $d$-open (equivalently $e$-open) means being open in the usual (standard) topology on $\Bbb R^2.$
(V). Let $H=\{(x,y)\in \Bbb R^2: xy\ne 1\}.$ For $p=(x,y)\in H$ let $xy=1+z.$ 
We have $z\ne 0$. Take  $s_p>0$ such that $s_p^2<|z|/3$ and $s_p|y|<|z|/3$ and $s_p|x|<|z|/3.$ 
For $(x',y')\in B_e(p,s_p)$ let $x'=x+u$ and $y'=y+v.$ We have $|u|<s_p$ and $|v|<s_p.\;$ And  $$x'y'-1=(x+u)(y+v)-1=(xy-1)+(uy+vx+uv)=z+(uy+vx+uv).$$ Now $$|uy+vx+uv|\leq |u|\cdot |y|+|v|\cdot |x|+|u|\cdot |v|\leq$$ $$\leq  s_p|y|+s_p|x|+s_p^2<|z|/3+|z|/3+|z|/3=|z|.$$ So $z\ne -(uy+vx+uv),$ implying $x'y'\ne 1,$ implying  $(x',y')\in H.$ Therefore $B_e(p,s_p)\subset H.$ So $H$ is open.
