Definition A function $f : X \to Y$ between two topological spaces is continuous at $x$ if for any $V(f(x))$ open set containing $f(x)$ there's $U(x)$ open containing $x$ such that $f(U(x)) \subset V(f(x))$.
Consider $[0,1]$ with the subspace topology induced from $\mathbb{R}$ with the usual topology. Defining the function
$$
f(x) = \begin{cases}
0 & 0 \leq x < 1/2 \\
1 & 1/2 \leq x \leq 1
\end{cases}
$$
I want to show that this function isn't continuous at $1/2$. I'm trying to learn how to apply the definition I stated.
Let $\epsilon > 0$ and set $V(1) = (1 – \epsilon, 1 + \epsilon)$ I'm considering two cases.
- $1 – \epsilon > 0$. This implies
$f^{-1}(V(1)) = [1/2,1]$ - $1 – \epsilon \leq 0$. This implies $f^{-1}(V(1)) = [0,1]$
This should cover all the cases, now from what I see here I have case 2) which is an open set in the induced topology, however 1) is not an open set. Therefore by 1) we can find an open sets in $Y$ such that there's no open set as pre-image, so $f$ is not continuous.
Am I applying the definition correctly?
Best Answer
You want to prove that $ f $ is Not continuous at $ \frac 12$, so you have to take the negation of your definition.
In other words, you should show that
$$\exists V(1)\;\;:\;\;\forall U(\frac 12)\;\; f(U(\frac 12))\not\subset V(1)$$
So, if we can take $$V(1)=(1-\frac 13,1+\frac 13)$$ and observe that
$$\forall \eta>0\;\; f((\frac 12-\eta, \frac 12+\eta))=\{0,1\}$$ and $$\{0,1\}\not\subset (\frac 23,\frac 43)$$