Can $\le$ be used insted of < in the definition of continuity?

Solution 1:

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.

Solution 2:

Suppose $f\colon X\rightarrow Y$ is continuous at $x\in X$ in the usual sense. That is, $$ \forall\epsilon>0\colon\exists\delta>0\colon\forall x^{\prime}\in X\colon d_{X}\left(x,x^{\prime}\right)<\delta\Rightarrow d_{Y}\left(f\left(x\right),f\left(x^{\prime}\right)\right)<\epsilon. $$ Let $\epsilon>0$. Then there exists $\delta$ s.t. for all $x^{\prime}\in X$, $d_{X}\left(x,x^{\prime}\right)<\delta$ implies $d_{Y}\left(f\left(x\right),f\left(x^{\prime}\right)\right)<\epsilon\leq\epsilon$. 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, $$ \forall\epsilon>0\colon\exists\delta>0\colon\forall x^{\prime}\in X\colon d_{X}\left(x,x^{\prime}\right)\leq\delta\Rightarrow d_{Y}\left(f\left(x\right),f\left(x^{\prime}\right)\right)\leq\epsilon. $$ Let $\beta>0$ and $\epsilon-\beta>0$. Then there exists $\delta$ s.t. for all $x^{\prime}\in X$, $d_{X}\left(x,x^{\prime}\right)<\delta\leq\delta$ implies $d_{X}\left(f\left(x\right),f\left(x^{\prime}\right)\right)\leq\epsilon-\beta<\epsilon$.