Example of closure of image not equals image of closure under continuous mapping

Question

Let $f:X\to Y$ be a continuous mapping, and let $E\subseteq X$, I have proved that $$ f(\overline{E})\subseteq \overline{f(E)} $$ But I got stuck when trying to find an example such that $f(\overline{E})$ is a proper subset of $\overline{f(E)}$, can anyone provide an idea?

Answer 1

Lets consider $f \colon \mathbb{R}\setminus\{0\} \to \mathbb{R}$ as $f(x) = 1/x$ where $\mathbb{R}\setminus\{0\}$ is taken with the subspace topology induced by the standar topology of $\mathbb{R}$.

Since $\mathbb{R}\setminus\{0\}$ is closed in $\mathbb{R}\setminus\{0\}$, we have that $f(\overline{\mathbb{R}\setminus\{0\}}) = f(\mathbb{R}\setminus\{0\}) = \mathbb{R}\setminus\{0\}$, but $\overline{f(\mathbb{R}\setminus\{0\})} = \mathbb{R}$.

Answer 2

Let $X=\mathbb Z_{> 0}$ and $Y= \mathbb R$. Now $X=\mathbb Z_{>0}$ is discrete, hence any function $f\colon X \to Y$ is continuous and any subset $I\subseteq X$ is closed. It is therefore easy to find examples of functions $f$ for which $\overline{f(X)}\supseteq f(X)= f(\bar{X})$. For example:

1. Let $f(n)= 1/n$. Then $\overline{f(X)} = \{0\}\cup f(X)\supsetneq f(X)$.
2. Let $f(n)=q_n$ where $q_1,q_2,\ldots,$ is an enumeration of $\mathbb Q \subseteq \mathbb R$. Then $\overline{f(X)} = \mathbb R\supsetneq f(X)=\mathbb Q$.

Answer 3

Consider $X = \{0, 1\}$ in the discrete topology, and $Y = \{0, 1\}$ in the indiscrete topology. Let $f : X \to Y$ be the identity function. Clearly, $f$ is continuous. Now take $E = \{0\} \subseteq X$. We have ${\rm Cl} E = E$ in $X$, and $f ({\rm Cl} E) = E$. But the closure of $f (E) = E$ in $Y$ is $Y$, and $E \subset Y$ holds strictly.