# Product topology questions

Pick out the true statements.

a. Let $\{X_i:i\in I\}$ be topological spaces. Then the product topology is the smallest topology on $X = \prod X_i$ such that each of the canonical projections $\pi : X \to X_i$ is continuous.

b. Let $X$ be a topological space and $W\subseteq X$. Then, the induced subspace topology on $W$ is the smallest topology such that $\mathrm{id}\upharpoonright W : W\to X$, where $\mathrm{id}$ is the identity map, is continuous.

c. Let $X =\Bbb R^n$ with the usual topology. This is the smallest topology such that all linear functionals on $X$ are continuous.

• What, if anything, have you tried? Also, on this site we'd prefer that you not use the imperative ("Pick out..."). It'd ruffle fewer feathers to begin with something like "I can't figure out which of these is true. For (a) I tried <your thoughts on the problem> but I got stuck at <some step> and I haven't got a clue about how to approach (b) and (c)." – Rick Decker Sep 23 '12 at 0:44

b. This is also just a matter of checking the definitions: if $\mathrm{id}\upharpoonright W:W\to X$ is continuous, then for every open set $U$ in $X$, $\left(\mathrm{id}\upharpoonright W\right)^{-1}[U]$ must be open in $W$. What is another description of the set $\left(\mathrm{id}\upharpoonright W\right)^{-1}[U]$, one that does not mention any function? Is it true that the collection of all of these inverse images is a topology on $W$?
c. The projections from $\Bbb R^n$ to $\Bbb R$ are linear functionals; use (a).
• For (a), the product topology is the smallest topology that contains all of the usual basic open sets. If continuity of the projections implies that the usual basic open sets of the product topology are open, then any topology making the projections continuous must contain the product topology. Conversely, the product topology does make the projections continuous. Therefore it is the smallest topology that does so. This also takes care of (c), and (b) is similar: the subspace topology is the smallest one for which all of the sets $U\cap W$ with $U$ open in $X$ are open in $W$, so ... – Brian M. Scott Sep 16 '12 at 6:48