# Thread: f is not continuous on R and discrete metric

1. ## f is not continuous on R and discrete metric

f:R->R
f(x) is 1 on Q and 0 elsewhere

Show f is not continuous at any point of R.
What about if the usual metric on R is replaced by the discrete metric?

I know that if f(x) takes 1 on Q and 0 elsewhere(Q complement), then by the density property there's infinite number of points between any values of Q, so the graph will keep jumping from 1 to 0 and back, so that's obviously not continuous. Have no idea how to show that in an actual proof though.

Also, I don't really understand replacing a metric by some other metric. What happens id R is replaced by the discrete metric?

2. what happens if you use de descrete metric is that any set is open(descrtete toplogy) so any function f:R->R will be continous.(look for definition of continuity in topological spaces)

3. Originally Posted by cp05
f:R->R
f(x) is 1 on Q and 0 elsewhere

Show f is not continuous at any point of R.
What about if the usual metric on R is replaced by the discrete metric?

I know that if f(x) takes 1 on Q and 0 elsewhere(Q complement), then by the density property there's infinite number of points between any values of Q, so the graph will keep jumping from 1 to 0 and back, so that's obviously not continuous. Have no idea how to show that in an actual proof though.

Also, I don't really understand replacing a metric by some other metric. What happens id R is replaced by the discrete metric?
I assume that $\displaystyle \mathbb{R}$ is intially under the usual topology. Then, suppose that $\displaystyle f$ is continuous at $\displaystyle x_0\in\mathbb{R}$. Then, since $\displaystyle \mathbb{Q},\mathbb{R}-\mathbb{Q}$ are both dense in $\displaystyle \mathbb{R}$ there exists sequences $\displaystyle \{q_n\},\{i_n\}$ in each such that $\displaystyle q_n,i_n\to x_0$. Thus, $\displaystyle 0=\lim\text{ }0=\lim\text{ }f(i_n)=f(x_0)=\lim\text{ }f(q_n)=\lim\text{ }1=1$. That looks strange, huh?

Also, if $\displaystyle X$ is a space with the discrete topology then any mapping $\displaystyle f:X\to Y$ is continuous regardless of $\displaystyle f$ or $\displaystyle Y$. This is clear since if $\displaystyle U\subseteq Y$ is open then $\displaystyle f^{-1}(U)\subseteq X$ but since every subset of $\displaystyle X$ is open we see in particular that $\displaystyle f^{-1}(U)$ is open.

4. Also, if $\displaystyle X$ is a space with the discrete topology then any mapping $\displaystyle f:X\to Y$ is continuous regardless of $\displaystyle f$ or $\displaystyle Y$. This is clear since if $\displaystyle U\subseteq Y$ is open then $\displaystyle f^{-1}(U)\subseteq X$ but since every subset of $\displaystyle X$ is open we see in particular that $\displaystyle f^{-1}(U)$ is open.[/quote]

How do I know that every subset of X is open? Thanks so much.

5. Originally Posted by cp05
Also, if $\displaystyle X$ is a space with the discrete topology then any mapping $\displaystyle f:X\to Y$ is continuous regardless of $\displaystyle f$ or $\displaystyle Y$. This is clear since if $\displaystyle U\subseteq Y$ is open then $\displaystyle f^{-1}(U)\subseteq X$ but since every subset of $\displaystyle X$ is open we see in particular that $\displaystyle f^{-1}(U)$ is open.
How do I know that every subset of X is open? Thanks so much.[/QUOTE]

That's what the discrete topology means. The discrete metric gives rise to the discrete topology. To see this merely let $\displaystyle X$ have the discrete metric and let $\displaystyle E\subseteq X$ be arbitrary. Then, given any $\displaystyle x\in X$ we have that $\displaystyle B_{\frac{1}{2}}(x)\cap E$ contains at most one point of $\displaystyle E$ and so $\displaystyle x\notin D(E)$ ($\displaystyle D(E)$ is the derived set...the set of limit points). Since $\displaystyle x\in X$ was arbitrary it follows that $\displaystyle D(E)=\varnothing$. Thus, trivially $\displaystyle D(E)\subseteq E$ and thus $\displaystyle E$ is closed....BUT! This implies that if $\displaystyle G\subseteq X$ then $\displaystyle G$ is open since $\displaystyle X-G$ is closed (since any subset of $\displaystyle X$ is closed)