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Normally continuity of a map between normed spaces is defined as :

A map $f : X \rightarrow Y$ is continuous at $x_0$, if for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x \in X$ satisfying $||x - x_0\|| < \delta $, we have $||f(x) - f(x_0)|| < \epsilon$.

I am wondering that if the following is equivalent:

A map $f : X \rightarrow Y$ is continuous at $x_0$, if for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x \in X$ satisfying $||x - x_0\|| \leq \delta $, we have $||f(x) - f(x_0)|| \leq \epsilon$.

Perhaps I am being overly analytic, but I would still like to know if there is any harm is using the latter definition.

RobPratt
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amitoz
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    No; the standard definition clearly implies yours, and if we take $\epsilon'=\epsilon/2$, then having $||f(x)-f(x_0)||\leq \epsilon'$ implies $||f(x)-f(x_0)||\lt\epsilon$. Similarly, if $\delta=\delta'/2$ then any $x$ with $||x-x_0||\leq\delta'$ will have $||x-x_0||\lt\delta$. – Steven Stadnicki Feb 07 '22 at 19:12
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    They are equivalent definitions, see e.g. Why does the definition of limits of a function have strict inequality? – Minus One-Twelfth Feb 07 '22 at 19:13
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    This is something that you will learn over time. Basically everything (except the order of quantifiers) in the definition of limit can be changed. $< 5000 \epsilon$ is good enough $\le \delta/2$ also gives an equivalent definition etc. – J. De Ro Feb 07 '22 at 20:16
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    If you have a look at how to interpret statements in the first-order language, you may see why they are equivalent. The question you gave is what I exactly had when I first learned the analysis. – Hermis14 Feb 07 '22 at 20:26

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