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Is being $T_1$ is a topological invariant? Is being a first-countable space is a topological invariant? I need a little hint as to whether or not these sets are topological invariants.

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    $\begingroup$ Suppose $X$ is T$_1$ and $f:X\to Y$ is a homeomorphism. Can you show $Y$ is T$_1$? Repeat for $X$ first countable. $\endgroup$ Dec 16, 2014 at 2:02
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    $\begingroup$ By "be first accounting space", do you mean "being a first-countable space"? If so, yes, first countability is a topological invariant. Here is a list: en.wikipedia.org/wiki/Topological_property $\endgroup$
    – Unit
    Dec 16, 2014 at 2:46

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A toplogical invariant is a property that is preserved under homeomorphism. So your first question is equivalent to:

If $X$ and $Y$ are homeomorphic, and $X$ is $T_1$, is $Y$ also $T_1$?

Let $f\colon Y \to X$ be a homeomorphism. Given that $X$ is $T_1$, and $f$ is continuous and invertible, can you show $Y$ is $T_1$?

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Let $X$ be $T_{1}$. Let $f:X\rightarrow Y$ be a homeomorphism. This tells us that $f$ is an open mapping (i.e., open sets map to open sets). Let $x,y ∈ X$, and let $B$ be an open set containing $x$ but not $y$. Since, $f$ is an open map, then $f(B)$ will be an open set in $Y$. Can we be certain that the open set $f(B)$ contains $f(x)$ but does not contain $f(y)$? If so, then it follows that $Y$ is also a $T_{1}$ space, and thus topologically invariant.

Secondly, regarding first-countable spaces, let $X$ be a first-countable space, and let $x$ be any element of $X$. Then $x$ has a countable neighborhood basis. Since $f$ is an open (closed) map, it will map every open (closed) neighborhood around $x$ (in $X$) to one around $f(x)$ (in $Y$). Thus, it follows that $f(x)$ has a countable neighborhood basis, and thus $Y$ is first-countable.

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