f is not continuous on R and discrete metric

**cp05** · April 6th 2010, 11:12 AM

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?

**facenian** · April 6th 2010, 11:26 AM

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)

**Drexel28** · April 6th 2010, 11:54 AM

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 $\mathbb{R}$ is intially under the usual topology. Then, suppose that $f$ is continuous at $x_0\in\mathbb{R}$ . Then, since $\mathbb{Q},\mathbb{R}-\mathbb{Q}$ are both dense in $\mathbb{R}$ there exists sequences $\{q_n\},\{i_n\}$ in each such that $q_n,i_n\to x_0$ . Thus, $0=\lim\text{ }0=\lim\text{ }f(i_n)=f(x_0)=\lim\text{ }f(q_n)=\lim\text{ }1=1$ . That looks strange, huh?

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

**cp05** · April 6th 2010, 12:13 PM

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

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

**Drexel28** · April 6th 2010, 12:18 PM

Originally Posted by cp05

Also, if $X$ is a space with the discrete topology then any mapping $f:X\to Y$ is continuous regardless of $f$ or $Y$ . This is clear since if $U\subseteq Y$ is open then $f^{-1}(U)\subseteq X$ but since every subset of $X$ is open we see in particular that $f^{-1}(U)$ 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 $X$ have the discrete metric and let $E\subseteq X$ be arbitrary. Then, given any $x\in X$ we have that $B_{\frac{1}{2}}(x)\cap E$ contains at most one point of $E$ and so $x\notin D(E)$ ( $D(E)$ is the derived set...the set of limit points). Since $x\in X$ was arbitrary it follows that $D(E)=\varnothing$ . Thus, trivially $D(E)\subseteq E$ and thus $E$ is closed....BUT! This implies that if $G\subseteq X$ then $G$ is open since $X-G$ is closed (since any subset of $X$ is closed)

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