### NBHM 2020 PART C Question 27 Solution ($f(x,y) = x + y$ is a clopen function, preserves dense and discrete sets)

Consider $f : \Bbb R^2 \to \Bbb R$ defined by $f(x,y) = x + y$. For each of the following statements, state whether it is true or false.
1. Image under $f$ of any open set if open,
2. Image under $f$ of any closed set is closed,
3. Image under $f$ of any dense set is dense,
4. Image under $f$ of any discrete set is discrete.

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Solution:
Option 1: (True) Consider the function $f$. We claim that $f$ is an open map. Let $U$ be an open set in $\Bbb R^2$. We need to show that $f(U)$ is open in $\Bbb R$. Let $a \in f(U)$, we will show that there is a neighborhood of $a$ which is entirely contained in $f(U)$. Let $(x,y) \in U \subseteq \Bbb R^2$ be such that $f(x,y)=x+y = a$. Since $a \in f(U)$ this is possible. Now $(x,y) \in U$ and $U$ is open and so we can find an $\epsilon > 0$ such that the open set $V = (x-\epsilon,x+\epsilon) \times (y-\epsilon,y+\epsilon) \subseteq U$. Now, $f(V) = ((x+y)-2\epsilon, (x+y)+2\epsilon) = (a-2 \epsilon, a+2\epsilon)$  is an open interval. Since $V \subset U$ we have $f(V) \subseteq f(U)$ is an interval containing $a$. Therefore $f(U)$ is open.
option 2:(False) We will construct a closed set $F$ such that $f(F)$ is not closed. Let $F = \Bbb Z \times \sqrt r \Bbb Z$ where $r$ is not a perfect square integer. Then $f(F) = \Bbb Z + \sqrt r \Bbb Z$ which is a dense subset of $\Bbb R$. (Why this is dense? please check my next post). Therefore $f(F)$ is not closed.
option 4:(False) Consider the example given in option 2. $F$ is discrete but $f(F)$ is not.
option 3: (True) Let $D$ be a subset of a metric space $X$, then $D$ is dense in $X$ if given $x$ in $X$ there exists a sequence $d_n$ in $D$ such that $d_n$ converges to $x$. We will use this definition. Consider the function $f$ and let $D$ be a dense subset of $\Bbb R^2$. We claim that $f(D)$ is dense in $\Bbb R$. Let $a \in f(D)$ then there exists a point $(x,y) \in D$ such that $f(x,y) = a$. Now $(x,y) \in D$ and $D$ is dense. So there exists a sequence $(x_n,y_n) \in D$ such that $(x_n,y_n)$ converges to $(x,y)$. Since the function $f$ is continuous, we have $f((x_n,y_n)) = (x_n+y_n)$ converges to $x+y = a$. Since $(x_n,y_n)$ is a sequence in $D$ we have $(x_n+y_n)$ is a sequence in $f(D)$ which converges to $a$. Therefore $f(D)$ is dense in $\Bbb R$.
THERE IS A MORE GENERAL RESULT IS TRUE:
Every continuous subjective (onto) map between the metric spaces preserves the dense sets. The proof is essentially the same as above. Note that, in this problem given $f$ is a surjective continuous map.
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1. I have doubt Sir
inverse image of open set is open.under continuous function

2. What is the doubt?

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