This is regarding the equivalent formulations of continuity in Tao's Analysis I, Proposition 9.4.7. I'm having trouble reconciling (c) and (d) below with the definition of convergence of functions in Definition 9.3.6. Proposition 9.4.7 states:
Let $X$ be a subset of $\mathbf{R}$, let $f:X\rightarrow \mathbf{R}$ be a function, and let $x_0$ be an element of $X$. Then the following four statements are logically equivalent:
(a) $f$ is continuous at $x_0$
(b) For every sequence $(a_n)$ consisting of elements of $X$ with $\lim_{n \to \infty}a_n = x_0$, we have $\lim_{n \to \infty}f(a_n) = f(x_0)$
(c) For every $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x)-f(x_0)|<\epsilon$ for all $x\in X$ with $|x-x_0|<\delta$
(d) For every $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x)-f(x_0)|\le\epsilon$ for all $x\in X$ with $|x-x_0|\le\delta$
However, in Definition 9.3.6, convergence of a function at a point is defined as:
we have $\lim_{x \to x_0; x \in E}f(x)=L$ iff for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x)-L|\le\epsilon$ for all $x\in E$ such that $|x-x_0|<\delta$.
The issue I'm having is that this definition uses $<$ for the $\delta$ but $\le$ for the $\epsilon$. Based on this, I would have guessed the corresponding definition of continuity to also involve $<$ for the $\delta$ part and $\le$ for the $\epsilon$ part. But what we have is two separate (equivalent) definitions, one using only $<$ and another using only $\le$. I think I'm missing the intuition here. Does this mean that the type of inequality doesn't matter? Would it be OK to also have another definition, say (e) as follows:
(e) For every $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x)-f(x_0)|\le\epsilon$ for all $x\in X$ with $|x-x_0|<\delta$
Any help will be greatly appreciated!
