# Continuity definiton problem

• September 26th 2008, 08:56 AM
Continuity definiton problem
Show that $f: A \rightarrow R^{m}$ is continuous at $x_0$ iff for every $\epsilon > 0$, there exist $\delta > 0$ such that $|| x-x_0 || \leq \delta$, we have $||f(x)-f(x_0)|| \leq \epsilon$

So I can replace the definition < with $\leq$?
• September 26th 2008, 10:08 AM
Plato
Quote:

Originally Posted by tttcomrader
Show that $f: A \rightarrow R^{m}$ is continuous at $x_0$ iff for every $\epsilon > 0$, there exist $\delta > 0$ such that $|| x-x_0 || \leq \delta$, we have $||f(x)-f(x_0)|| \leq \epsilon$
So I can replace the definition < with $\leq$?

I am not sure that I follow your question.
Do note that the tradition definition is for “<”.
However, it is also $\left( {\forall \varepsilon > 0} \right)\left[ {\frac{\varepsilon }
{2} < \varepsilon } \right]$
, that is for all.
So $\left\| {f(x) - f(x_0 )} \right\| \leqslant \frac{\varepsilon }
{2} \Rightarrow \quad \left\| {f(x) - f(x_0 )} \right\| < \varepsilon$
• September 27th 2008, 09:26 AM
Well, from the last post I understand how to go from $\leq$ to <