Can $\le$ be used insted of < in the definition of continuity? A common definition of a continuous map $T:M_1\to M_2$ is that for every $x\in M_1$ and every $\varepsilon>0$ there exists a $\delta >0$ such that for all $y$ in $M_1$
$$d_1(x,y)<\delta \implies d_2(T(x),T(y))<\varepsilon,$$
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 $\varepsilon>0$ there is a $\delta>0$ such that
$$\|Tx-Ty\|\le\varepsilon\ \mbox{for all $y$ in $M_1$ satisfying } \|x-y\|\le \delta.$$
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!
 A: 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.
A: Suppose $f\colon X\rightarrow Y$ is continuous at $x\in X$ in the
usual sense.
That is,
$$
\forall\varepsilon>0\colon\exists\delta>0\colon\forall x^{\prime}\in X\colon d_{X}\left(x,x^{\prime}\right)<\delta\Rightarrow d_{Y}\left(f\left(x\right),f\left(x^{\prime}\right)\right)<\varepsilon.
$$
Let $\varepsilon>0$. Then there exists $\delta$ s.t. for all $x^{\prime}\in X$,
$d_{X}\left(x,x^{\prime}\right)<\delta$ implies $d_{Y}\left(f\left(x\right),f\left(x^{\prime}\right)\right)<\varepsilon\leq\varepsilon$.
Take $\delta^{\prime}\equiv\delta-\alpha$ for some $0<\alpha<\delta$.
Suppose now that $f$ is continuous at $x$ in the ``exotic'' sense.
That is,
$$
\forall\varepsilon>0\colon\exists\delta>0\colon\forall x^{\prime}\in X\colon d_{X}\left(x,x^{\prime}\right)\leq\delta\Rightarrow d_{Y}\left(f\left(x\right),f\left(x^{\prime}\right)\right)\leq\varepsilon.
$$
Let $\beta>0$ and $\varepsilon-\beta>0$. Then there exists $\delta$
s.t. for all $x^{\prime}\in X$, $d_{X}\left(x,x^{\prime}\right)<\delta\leq\delta$
implies $d_{X}\left(f\left(x\right),f\left(x^{\prime}\right)\right)\leq\varepsilon-\beta<\varepsilon$.
A: As it is said in the previous answers, you can use either the definition with all inequalities strict ($\lt$) or either the definition with all inequalities "less or equal" ($\leq$).
What you cannot do is using $\leq$ for the $\delta$ and the $\lt$ for $\epsilon$ (you cannot either use $\lt$ for $\delta$ and $\leq$ for $\epsilon$).
The reason is, as janmarqz said, that we have continuity of a function if the preimage of every closed set is a closed set OR if the preimage of every open set is an open.
One can not mix the two...
