# function is continuous iff its composition with a curve is continuous

$f: R^n \to R \space is \space continuous \space \iff \space \forall \space \gamma: [a,b] \to R^n \space . \space f \circ \gamma : [a,b] \to R \space is \space continuous$

I succeeded to prove the first direction(given f is continuous). This seems to be the "easy" direction. As for the other direction, I thought about the following solution but I'm not quite sure it is correct:

let $\epsilon$ be a positive number. We know that $f\circ \gamma$ is continuous, thus if $\forall t_1 \in [a,b] \space. \exists \delta>0 .\forall t_2 \in [a,b] . |\gamma(t_1)-\gamma(t_2)| <\delta => \space |f(\gamma(t_1)) - f(\gamma(t_2))|<\epsilon$

In addition , w also know that $\gamma$ is a curve and therefore a continuous function: $\forall t_1 \in [a,b] \space. \exists \delta _1 >0 .\forall t_2 \in [a,b] . |t_1-t_2| <\delta_1 => \space |\gamma(t_1) - \gamma(t_2)|<\delta$

now, for conclusion : $\forall t_1 \in [a,b] \space. \exists \delta _1 >0 .\forall t_2 \in [a,b] . |t_1-t_2| <\delta_1 => \space |\gamma(t_1) - \gamma(t_2)|<\delta =>|f(\gamma(t_1)) - f(\gamma(t_2))|<\epsilon$

and this is correct for every $t_1,t_2 \in R$ and for every $\gamma(t_1),\gamma(t_2) \in R^n$ and therefore f is continuous. $\blacksquare$

I'm quite sure it is not enough , so please help me solve it correctly or tell me if this is a good solution.

Thanks.

The proposal fails because once a choice of $\gamma$ is made, too many points are missing. Let's try by way of contradiction. Suppose $f$ is not continuous at $x_0$. This means there is $\varepsilon>0$ such that for every $k\ge1$ there is a point $x_k$ such that $\|x_k-x_0\|<1/k$ but $\|f(x_k)-f(x_0)\|>\varepsilon$. What we need is a curve $\gamma$ from $x_1$ to $x_0$ passing through all $x_k$. This can be produced glueing linear interpolations $\gamma_k:[1-1/k,1-1/(k+1)]\to[x_k,x_{k+1}]$. All together define a continuous function $\gamma:[0,1)\to{\mathbb R}^n$ that extends continuously by $\gamma(1)=x_0$ by construction. Also by construction $f\circ\gamma$ is not continuous at $1$.