We have the standard topological definition of continuity as:

For topological spaces, $(X,\tau_x)$ and $(Y,\tau_y)$, a function $f : X \rightarrow Y$ is continuous if $\forall V \in \tau_Y$, $f^{-1}(V) \in \tau_X$.

I was presented with a similar property, namely:

Let $f : X \rightarrow Y$, where $\forall U \subseteq X$, $f(U) \in \tau_Y$ implies $U \in \tau_X$

How is this different from the definition of continuity? It seems to me that the property above starts from the domain into the range, while continuity is the reverse. Perhaps, could anyone show me examples of $f$ that is continuous but the property does not hold, and vice versa?

  • 1
    $\begingroup$ @happymath The second property is not the definition of an open map. $\endgroup$ Mar 12, 2017 at 16:50
  • $\begingroup$ @SeverinSchraven Yes you are right. I read it wrong. Thanks for correcting. $\endgroup$
    – happymath
    Mar 12, 2017 at 16:51

2 Answers 2


Let $Y$ be a singleton.

Then every function $f:X\to Y$ is continuous (no matter what topology we have on $X$).

However the existence of a function $f:X\to Y$ that has the second property implies that $X$ is equipped with the discrete topology. If so then moreover every function $X\to Y$ will have the second property.


Let $X=\{1,2\}$, $Y=\{0,1,2\}$ and let $f:X\to Y$ be prescribed by $x\mapsto x$.

Let $\tau_X=\{\varnothing,\{1,2\}\}$ and let $\tau_Y=\{\varnothing,\{0,1\},\{0,1,2\}\}$.

Then $f$ is not continuous, because $f^{-1}(\{0,1\})=\{1\}$ is not open in $X$.

On the other hand $f$ satisfies the second property.

  • $\begingroup$ Could you elaborate as to why the $Y$ being a singleton implies that every function is continuous? $\endgroup$
    – H. Moon
    Mar 12, 2017 at 17:07
  • $\begingroup$ And do you mean the existence of a continuous function with teh second property implies that $X$ will have the discrete topology? $\endgroup$
    – H. Moon
    Mar 12, 2017 at 17:08
  • $\begingroup$ If $Y$ is a singleton then it must be equipped with indiscrete topology: $\tau_Y=\{\varnothing,Y\}$ (so any function to $Y$ is continuous). On your second comment: suppose such $f$ exists and that $U$ is a subset of $X$. Then $f(U)=\varnothing\in\tau_Y$ or $f(U)=Y\in\tau_Y$ so in both cases we are allowed to conclude that $U\in\tau_X$. This for every subset of $X$. $\endgroup$
    – drhab
    Mar 12, 2017 at 17:13

Let $\vert \cdot \vert$ be the standard norm on $\mathbb{R}$. We denote by $(\mathbb{R}, \vert \cdot \vert)$ the topological space, where the topology is induced by the norm.

Let $f: (\mathbb{R}, \{ \emptyset, \mathbb{R} \}) \rightarrow (\mathbb{R}, \vert \cdot \vert), \ f(x)=sgn(x)$ is not continuous, but it fulfills the second property (as $f(U)$ is open in the norm topology iff $U=\emptyset$). Note that this map is not open, as $f(\mathbb{R})=\{-1, 0, 1 \}$ is not open in the norm topology (i.e. the second property does not coincide with being an open map).

On the other hand, the map $g: (\mathbb{R}, \vert \cdot \vert) \rightarrow (\mathbb{R}, \vert \cdot \vert), \ g(x)=x^2$ is continuous, but it does not fulfill the second property. Namely

$$ g([-1,0) \cup [1, 2))=(0,4) $$

is open, but $ [-1,0) \cup [1, 2)$ isn't.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.