Closed Intervals How do topologists prove continuity of a function with the usual topology at the endpoints of a closed interval?  For instance, how would a topologist prove continuity for $f(x)=x^2$ on the closed interval $[0,1]$ at the point $x = 1$ using open sets (not using calculus)?
 A: It is simply a matter of translating the $\varepsilon$-$\delta$ definition to a metric language. Saying that given any $\varepsilon >0$ there exists a $\delta >0$ such that $|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon$ is saying that, given any open ball $V=B(f(a),\varepsilon)$ there is an open ball $U=B(a,\delta)$ such that $f(U)\subseteq V$. One then uses that open balls are a basis for the open sets in any metric space. This matches the general definition, that a map $X\to Y$ is open iff the preimage of any open set in $Y$ is open in $X$. Indeed, suppose that $U$ is open in $Y$, and consider $f^{-1}U$. Pick a point $a\in f^{-1}U$. This means $f(a)\in U$. Since $U$ is open, there is a ball $B(f(a),\varepsilon)\subseteq U$. By definition, there exists a ball $B(a,\delta)$ such that $f(B(a,\delta))\subseteq B(f(a),\varepsilon)\subseteq U$, so $B(a,\delta)\subseteq f^{-1}U$, i.e. $f^{-1}U$ is open. I leave the other direction to you.
A: Same way you would do it on an open interval. You have to show that inverse images of open sets are open, for $f(x)=x$ the inverse image of any set is the set itself, so it's true. It is only slightly harder for $f(x)=x^2$, but the inverse image of $(a,1]$ say, is $(\sqrt{a},1]$ and both are open (it's enough to consider intervals containing $1$ for continuity at $1$).
I think your issue is that open neighborhoods of $1$ do not "look" open on $[0,1]$ in the usual sense, but intersection of any open set in $\mathbb{R}$ with $[0,1]$ is nonetheless open in $[0,1]$ by definition of induced topology on $[0,1]$. So for example, $(\frac1n,1]$ are all open for any $n\in\mathbb{N}$.
