TITLE: Can $\le$ be used insted of < in the definition of continuity?
QUESTION [7 upvotes]: A common definition of a continuous map $T:M_1\to M_2$ is that for every $x\in M_1$ and every $\epsilon>0$ there exists a $\delta >0$ such that for all $y$ in $M_1$
$$d_1(x,y)<\delta \implies d_2(T(x),T(y))<\epsilon,$$
i.e. we use a strict inequality. Now in a proof it reads that $T$ is continuous if for every $x\in M_1$ and every $\epsilon>0$ there is a $\delta>0$ such that
$$\|Tx-Ty\|\le\epsilon\ \mbox{for all $y$ in $M_1$ satisfying } \|x-y\|\le \delta.$$ 
Here the norm is given by the metric. Is it correct to use $\le$ instead of the strict inequality <, and can one somehow prove the equality of these definitions? Or is this obviously the same condition?
Thanks in advance!

REPLY [7 votes]: Let $X$ and $Y$ two topological spaces then for the definition of continuity
there is an equivalence:
A map $f: X\to Y$ is continuous if and only if for each closed set $A$ in $Y$ we have $f^{-1}A$ is closed in $X$.
So, an answer to your question is included.