Which of the following sets are dense in $\Bbb R^2$ with respect to the usual topology.

1) $\{(x,y) \in \Bbb R^2 : x \in \Bbb N\}$,

2) $\{(x,y)\in \Bbb R^2 : x+y \text{ is rational}\}$,

3) $\{(x,y)\in \Bbb R^2 : x+y^2 = 5\}$,

4) $\{(x,y) \in \Bbb R^2 : xy \ne 0\}$.

A subset $A$ of a metric space $(X,d)$ is said to be dense if $A$ intersects all the non-empty open balls in $X$. In particular, to show a subset is not dense in $X$ enough to construct a non-empty open ball $U$ which is disjoint from $A$. i.e., $A \cap U = \Phi$.

Proof: Let $U$ be a non-empty open set in $X$. Since $A$ is dense in $X$, $A \cap U \ne \Phi$. This implies that $B \cap U \ne \Phi$ as $A \subseteq B$.

We have $\Bbb Q \times \Bbb Q$ is dense in $\Bbb R^2$ since the product of dense sets is dense. This is immediate using the projection maps. We have $\Bbb Q \times \Bbb Q \subseteq \{(x,y)\in \Bbb R^2 : x+y \text{ is rational}\}$ and hence the given set is dense in $\Bbb R^2$ by the above observation.

Case 1: $(a,b) \in D$, then clearly $B(x,r) \cap D \ne \Phi$.

Case 2: $(a,b) \notin D$. then $a = 0$ or $b=0$.

Hence the point $(a+\frac{r}{2},b+\frac{r}{2}) \in B(x,r)$.

Click 1) $\{(x,y) \in \Bbb R^2 : x \in \Bbb N\}$,

2) $\{(x,y)\in \Bbb R^2 : x+y \text{ is rational}\}$,

3) $\{(x,y)\in \Bbb R^2 : x+y^2 = 5\}$,

4) $\{(x,y) \in \Bbb R^2 : xy \ne 0\}$.

**Solution**:A subset $A$ of a metric space $(X,d)$ is said to be dense if $A$ intersects all the non-empty open balls in $X$. In particular, to show a subset is not dense in $X$ enough to construct a non-empty open ball $U$ which is disjoint from $A$. i.e., $A \cap U = \Phi$.

**option 1**:(**Not dense**) We claim that the given set $A := \{(x,y) \in \Bbb R^2 : x \in \Bbb N\}$ is closed. Hence it cannot be dense. Let $(x,y)$ be a limit point of this set. Then there exists a sequence $(x_n,y_n) \in A$ such that $(x_n,y_n)$ converges to $(x,y)$. This means that $x_n \to x$ and $y_n \to y$. Note that $(x_n,y_n) \in A$ implies that $x_n$ is a converging sequence of natural numbers. This is possible, only if, $x_n = k \in \Bbb N$ for sufficiently large $n$. In particular, this sequence is eventually constant and hence is converging to $k$. This shows that $x=k$ as the limit of a sequence is unique. Therefore $(x,y) \in A$ and $A$ is closed.**option 2**:(**True**) From the definition, we observe that, if $A \subseteq B$ and $A$ is dense in $X$, then $B$ is also dense in $X$.Proof: Let $U$ be a non-empty open set in $X$. Since $A$ is dense in $X$, $A \cap U \ne \Phi$. This implies that $B \cap U \ne \Phi$ as $A \subseteq B$.

We have $\Bbb Q \times \Bbb Q$ is dense in $\Bbb R^2$ since the product of dense sets is dense. This is immediate using the projection maps. We have $\Bbb Q \times \Bbb Q \subseteq \{(x,y)\in \Bbb R^2 : x+y \text{ is rational}\}$ and hence the given set is dense in $\Bbb R^2$ by the above observation.

**option 3**:(**False**) Again, we will prove that the set $C:= \{(x,y)\in \Bbb R^2 : x+y^2 = 5\}$ is closed in $\Bbb R^2$. Let $f: \Bbb R^2 \to \Bbb R$ be a function defined by $f(x,y) = x+y^2-5$, then being the polynomial function $f$ is continuous. Now $f^{-1}(0) = C$. Since $\{0\}$ is closed and $f$ is continuous, we have $C$ is closed.**option 4**:(**True**) Let $D: = \{(x,y) \in \Bbb R^2 : xy \ne 0\}$. We will prove that this set is dense in $\Bbb R^2$. Let $B(x,r)$ be a open ball in $\Bbb R^2$. We claim that $B((a,b),r) \cap D \ne \Phi$.Case 1: $(a,b) \in D$, then clearly $B(x,r) \cap D \ne \Phi$.

Case 2: $(a,b) \notin D$. then $a = 0$ or $b=0$.

**Subcase 1**: $a = 0$ and $b \ne 0$, then $(a,b)$ lies in the y axis. Consider the point $(a+\frac{r}{2},b) \in D$, then $d((a+\frac{r}{2},b),(a,b)) = \sqrt {|(a+\frac{r}{2}-a)^2 + (b-b)^2| } = \sqrt{\frac{r^2}{4}} = \frac{r}{2} < r$. Hence the point $(a+\frac{r}{2},b) \in B(x,r)$.**Subcase 2**: $a \ne 0$ and $b = 0$. Similar to the previous subcase consider the point $(a,b+\frac{r}{2}) \in D$.**Subcase 3**: $a=0$ and $b=0$. Consider the point $(a+\frac{r}{2},b+\frac{r}{2}) \in D$, then $d((a+\frac{r}{2},b+\frac{r}{2}),(a,b)) = \sqrt {|(a+\frac{r}{2}-a)^2 + (b+\frac{r}{2}-b)^2| } \\ = \sqrt{\frac{r^2}{4}+\frac{r^2}{4}} = \frac{\sqrt 2 r}{2} < r$Hence the point $(a+\frac{r}{2},b+\frac{r}{2}) \in B(x,r)$.

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