0
$\begingroup$

prove or disprove

Let $(X, \mathscr{T}_1)$ and $(Y, \mathscr{T}_2)$ be topological spaces and suppose that $f : X \to Y$ is a function that is $\mathscr{T}_1$ − $\mathscr{T}_2$ continuous. If f is one to one and $\mathscr{T}_2$ is the discrete topology on Y then $\mathscr{T}_1$ IS THE DISCRETE TOPOLOGY ON X.

I think,it is False statement since f is one to one but I do not know how can I come with counterexample.

$\endgroup$
  • $\begingroup$ The discrete topology can be characterised as the topology in which every point is an open set. Can you see how to prove it now? $\endgroup$ – Joppy Apr 27 '17 at 2:50
2
$\begingroup$

Take a point $y \in I_{2}$, $y$ is open by the definition of discrete topology.

So, if $f$ is continuos, then $f^{1}(y) = {x}$ is an open set of $I_{1}$.

Now use $f$ is one to one.

$\endgroup$
  • $\begingroup$ I know f is one to one that f(x1)=f(x2) implies x1=x2 but I do not know how can I apply it here could you help me please $\endgroup$ – rian asd Apr 27 '17 at 4:03
  • $\begingroup$ @rianasd: What is the preimage of the set $\{f(x)\}$? $\endgroup$ – celtschk Apr 27 '17 at 4:15
  • $\begingroup$ is it {y} ?.... $\endgroup$ – rian asd Apr 27 '17 at 4:17
  • $\begingroup$ @rianasd: No, the preimage is a subset of $X$. $\endgroup$ – celtschk Apr 27 '17 at 4:18
  • $\begingroup$ sorry, I am not sure what is it $\endgroup$ – rian asd Apr 27 '17 at 4:20
1
$\begingroup$

Suppose that $x \in X$. Then

$$f^{-1}[\{f(x)\}] = \{p \in X: f(p) \in \{f(x)\} \} = \{p \in X: f(p) =f(x) \} = \{x\}$$ where the last equality is a restatement that $f$ is 1-1.

$\{f(x)\}$ is open in $(Y,\mathscr{T}_2)$, as it is discrete. So it's inverse image under the continuous $f$ lies in $\mathscr{T}_1$, hence $\{x\}$ is open. This holds for every $x \in X$. Conclude that $X$ is indeed discrete.

$\endgroup$
1
$\begingroup$

Let $U$ be a subset of $x$. Thus $f[U]$ is an open subset of $Y$.
As $f$ is injective $U = f^{-1}[f[U]]$ is open.

$\endgroup$

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.