Show that if $f,g$ are continuous then so is $f\times g$

QN1: Consider the lower limit topology on $\mathbb{R}$ and the topology different from the lower limit topology given by the basis $\mathcal{B}=\{[a,b)|a < b,\, a\text{ and }b\text{ rational}\}$. Determine the closures of intervals $A=(0,\sqrt{2})$ and $B=(\sqrt{2},3)$ in these topologies.

QN2: Let $f\colon A \rightarrow B$ and $g\colon C \rightarrow D$ be continuous functions. Define a map $f\times g\colon A\times C \rightarrow B\times D$ by $(f\times g)(a\times c)=f(a)\times g(c)$.Show that $f\times g$ is continuous.

I think that the closure of $A$ in the lower limit topology is $[0,\sqrt{2})$ and the closure of $B$ in the lower limit topology is $[\sqrt{2},3)$. I would like to know the closures in the other topology.

For QN2 I think to attempt as follows.

By continuity of $f$ and $g$, the inverse images of $B$ and $D$ are open in $A$ and $C$, respectively. Then the product of these images should be open in $A\times C$. This is the way, I think, to deduce the continuity of $f\times g$, am I right?

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B={[a,b)|a<b, a and b rational –  aloyce temba Jun 21 '11 at 10:17
Please consider editting that in, rather than leaving it as a comment. Also, the reason that it is not showing up is because you are using the less-than and greater-than signs, which sometimes confuses the parser (thinking that it is HTML code). One workaround is to replace the less-than sign with the HTML entity &lt;, which shows as <. Similarly &gt; is >. –  Willie Wong Jun 21 '11 at 10:33
Simply having a space on either side of '<' helps. –  ShreevatsaR Jun 21 '11 at 11:38
Is this homework? If so, please add the (homework) tag. Otherwise, my apologies. –  Yuval Filmus Jun 21 '11 at 11:44

Question 1:

(a) [Lower Limit Topology] You are correct that the closures of $(0,\sqrt{2})$ and $(\sqrt{2},3)$ in the lower limit topology are $[0,\sqrt{2})$ and $[\sqrt{2},3)$, respectively.

(b) [Other Topology] Hint: Prove that the closure of $(0,\sqrt{2})$ in this topology is $[0,\sqrt{2}]$; use the irrationality of $\sqrt{2}$. Can you see that the closure of $(\sqrt{2},3)$ in this topology is $[\sqrt{2},3)$?

Question 2:

Let us choose the "standard basis" for the product topology on $B\times D$ consisting of all sets of the form $U\times V$ where $U$ is open in $B$ and $V$ is open in $D$. We wish to establish the continuity of $f\times g$.

(a) Show that it suffices to prove the preimage under $f\times g$ of each element of this basis is open in $A\times C$. (Hint: use the definition of a basis.)

(b) Prove that if $U$ is open in $B$ and $V$ is open in $D$, then $(f\times g)^{-1}(U\times V)=f^{-1}(U)\times g^{-1}(V)$. Deduce that $f\times g$ is a continuous function.

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As for your (Q2), by the universal property of the product of topological spaces, a map $\phi : X \longrightarrow B \times D$ is continuous if and only if its compositions $\pi_B \circ \phi$ and $\pi_D \circ \phi$ with each one of the projections $\pi_B : B\times D \longrightarrow B$ and $\pi_D : B \times D\longrightarrow D$ are.

Otherwise said: a map into a product $\phi : X \longrightarrow B \times D$, $\phi (x) = (\phi_B (x), \phi_D (x))$, is continuous if and only if all of its "components" $\phi_B = \pi_B \circ \phi : X \longrightarrow B$ and $\phi_D = \pi_D \circ \phi : X \longrightarrow D$ are.

But in our case, $\pi_B \circ (f\times g) = f \circ \pi_A$ and $\pi_D \circ (f\times g) = g \circ \pi_C$, where $\pi_A : A\times C \longrightarrow A$ and $\pi_C : A \times C \longrightarrow C$ are the projections, and these maps $f \circ \pi_A$ and $g \circ \pi_C$ are certainly continuous, since they are compositions of continuous maps.

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