# Continuity definiton problem

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

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

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

Do note that the tradition definition is for “<”.
However, it is also $\displaystyle \left( {\forall \varepsilon > 0} \right)\left[ {\frac{\varepsilon } {2} < \varepsilon } \right]$, that is for all.
So $\displaystyle \left\| {f(x) - f(x_0 )} \right\| \leqslant \frac{\varepsilon } {2} \Rightarrow \quad \left\| {f(x) - f(x_0 )} \right\| < \varepsilon$
• Sep 27th 2008, 09:26 AM
I asked the professor today and he said this problem is suppose to show that the two definitions for continuity are equiv.

Well, from the last post I understand how to go from $\displaystyle \leq$ to <

How to do this the other way around is killing me...

Any hints? Thanks.