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Let $X,Y$ be topological spaces and $f : X \rightarrow Y$ be a continous,surjective map.

Then which of the following are true -

$1)$ if $X$ is $2nd$ countable then $Y$ is $2nd$ countable.

$2)$ if $X$ is first countable then $Y$ is first countable.

$3)$if $X$ is $2nd$ countable then $Y$ is first countable.

$4)$if $X$ is First countable then $Y$ is $2nd$ countable.

My attempts : i know that every 2nd countable is first countable but every first countable need not be 2nd countable

here im confused that how can i solved this

Any hints/solution will be appreciated

thanks u

bof
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jasmine
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1 Answers1

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There are (lots of) examples of countable spaces $Y$ that are neither first- nor second-countable. Pick your favourite and fix it for now.

(Besides Arens-Fort space and the maximal compact topology, I gave constructions and proofs here, e.g. use $D(\omega)$ defined there or $(\omega, \mathcal{F}\cup\{\emptyset\})$ or $X(\mathcal{F})$ for $\mathcal{F}$ a free ultrafilter on $\omega$), which all have weight $\mathfrak{c}$)

Then let $X$ be the space with the set of $Y$ but in the discrete topology. $X$ is thus both second- and first-countable (singletons form a (local) base). Let $f$ be the identity, which is continuous as all maps from a discrete space are, and a surjection.

This is a counterexample to 1,2, 3 and 4. So all are false.

PatrickR
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Henno Brandsma
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