The following is from Wikipedia:
A locally convex space is defined to be a vector space $V$ along with a family of seminorms $\{p_α\}_{α ∈ A}$ on $V$.
A locally convex space carries a natural topology induced by the seminorms. By definition, it is the coarsest topology for which all the mappings $$ {\begin{cases}p_{\alpha ,y}:V\to \mathbf {R} \\x\mapsto p_{\alpha }(x-y)&y\in V,\alpha \in A\end{cases}}$$ are continuous. A base of neighborhoods of $y$ for this topology is obtained in the following way: for every finite subset $B$ of $A$ and every $ε > 0$, let $$ U_{B,\varepsilon }(y)=\{x\in V:p_{\alpha }(x-y)<\varepsilon \ \forall \alpha \in B\}. $$
Denote this topology as $\tau$. The function $f:=p_{\alpha,0}:V\to\mathbf{R}$ is continuous with respect to $\tau$. Could anyone explain why $$ f^{-1}(C)\in\tau $$ where $C=(1,2)$? ([Edit: Thanks to @silvascientist, I just asked a silly question. What I really want to know is how $f^{-1}(C)$ contains a basic set defined above.])
[Some thoughts:] When $C=(-\infty,r)$ for some $r\in\mathbf{R}$, I can see that $f^{-1}(C)\in\tau$. But I don't see how this might help.
Demanding that the seminorms $p_\alpha = p_{\alpha,0}$ be continuous provides us with a basis of neighborhoods of $0$, namely, the open semiballs $U_{\alpha, \varepsilon}(0)$, which are preimages of the open intervals $(-\varepsilon,\varepsilon)$ under the seminorms $p_\alpha$, and finite intersections thereof. (Although this doesn't completely ensure continuity of $p_\alpha$ since we still have to take pre-images of other basis sets in $\mathbb R$, not just basis neighborhoods of $0$, but we'll get to that in just a moment.) We can then turn this into a topology compatible with the vector space structure by translating neighborhoods of $0$ to arbitrary points, i.e, we obtain a neighborhood of $y$ by taking $U_{\alpha,\varepsilon}(0) + y = U_{\alpha,\varepsilon}(y)$. (Now we have that $p_\alpha$ is continuous - if we take any point $x \in V$, then for any $\varepsilon > 0$ we can take $U = U_{\alpha,\varepsilon}(x)$ to be a neighborhood of $x$ such that $p_\alpha(U) \subseteq (p_\alpha(x) - \varepsilon, p_\alpha(x) + \varepsilon)$. This also ensures that all the $p_{\alpha,y}$ are continuous, but I'll leave that as an exercise to the reader.
Now you're basis sets $U_{B,\varepsilon}(y)$ can be seen to be precisely the finite intersections of $U_{\alpha,\varepsilon}(y)$ mentioned above, and so are the basis we seek. This demonstrates that any open set in $V$ must then contain one of these sets, in particular the set $f^{-1}((1,2))$ you mention above, which must be open by continuity of $f$.