For the function
\begin{align*} f&:\mathbb{R}\to\mathbb{R},\\[5pt] f(&x) = 2x, \end{align*}
I know it is a continuous function in terms of the epsilon-delta definition, but how can I show it is continuous in terms of the topological definition of continuity (one of the equivalent definition of continuity)?
On
Nice question! It is always good to try and prove something multiple ways with multiple definitions just for the sake of exercise.
Let us use the most common (in my opinion) topological definition of continuity "$f \colon X \to Y$ is continuous if and only if $U$ open in $Y$ implies that $f^{-1}(U)$ is open in $X$.
In our case, the domain and codomain are both $\mathbb{R}$, and I assume you wish to consider them both with the Euclidean topology. We know that a basis for this topology is the collection of all open intervals $(a,b)$ where $a<b$. If you don't know this, look up topological basis. We also know it suffices to check continuity on a basis, if you don't it is an easy proof. So let $(a,b)$ be any such open set in the codomain. Then the preimage is $f^{-1}(a,b) = (a/2,b/2)$, which is still an open set.
Also, here is an interesting side-thought. If you know the proof that $f(x) = 2x$ is continuous via the epsilon delta definition, and you know the proof that epsilon delta continuity implies topological continuity (when the topological space is $\mathbb{R}^n$), you can actually use those two proofs to construct the proof you are looking for, since the latter is a constructive proof.
On
Write down the inverse of the function $f$: it is given by $f^{-1}(x) = \frac{x}{2}$.
Let $V \subseteq \mathbb{R}$ be open in $\mathbb{R}$, which means that $V = (a, b)$. We then need to show that $f^{-1}(V)$ is also an open set in $\mathbb{R}$.
$f^{-1}((a, b))= (\frac{a}{2}, \frac{b}{2})$ is clearly open in $\mathbb{R}$ which means that $f$ is a continuous function.
It suffices to check on a basis of the standard topology, so check the preimages of open intervals. Are they open?