This might be trivial, but I just want to make sure I got this right:

Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J \subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map

$F: X^I \to X^J$ given by $F((x_i)_{i \in I}) := (x_j)_{j \in J}$, i.e. $F$ is simply the projection from $I$ to $J$.

Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.

Thankful for any answers on this!


Of course this map is continuous: if $\pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $\pi_i$ ($i$ in its index set), i.e. $ \pi_i \circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $\pi_i \circ F = \pi_i$ for all $i \in J$ trivially, so $F$ is continuous.

  • $\begingroup$ You are right. Thanks a lot! $\endgroup$ – Marco Mar 28 '19 at 9:59

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