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Is the following statement true or false?

Let $f : X \rightarrow Y$ be a nonconstant continuous map of topological spaces. If $X$ is Hausdorff then $f(X)$ is Hausdorff.

My attempt: I was reading this answer but i did not understands anything enter image description here

can anybody explain this statement: If $X$ is Hausdorff then $f(X)$ is Hausdorff...... in detail and please tell me the solution..i would be more thankful.

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  • $\begingroup$ this is not true in general.. the author (Alireza) assumes that $Y$ is a $T_B$ space... and that $X$ is compact. $\endgroup$ – Yanko Oct 26 '17 at 15:40
  • $\begingroup$ @James : That's not necessarily true in general. The statement true if $f : X \rightarrow Y$ is a homeomorhism. $\endgroup$ – Sou Oct 26 '17 at 16:13
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A topological space $(X, \tau_X)$ is Hausdorff whenever every pair of distinct points have disjoint neighbourhoods. Note that this is really a condition on the topology $\tau_X$.

Take $f: X \to Y$. Define

$$ f(X) = \{ f(x) : x \in X \} \subseteq Y. $$

Whenever $f(X) = Y$ (that is, $f$ is onto), then it is clear what the topology on $f(X)$ is. But if $f(X)$ is a proper subset of $Y$, then we normally take the topology on $f(X)$ to be the relative topology

$$ \tau_f = \{ U \cap f(X) : U \in \tau_Y \}. $$

The claim (which, as noted in the comments, is not true in general) that $f(X)$ is Hausdorff whenever $X$ is Hausdorff should now be fairly self-explanatory.

To see why it is not true (even for continuous $f$), let $\tau_Y$ be the trivial topology consisting of only $Y$ and the empty set. $f$ is (trivially) continuous. Observe now that, as long as $Y$ is non-empty and not a singleton (and $f$ is not constant), $f(X)$ is not Hausdorff, even if $X$ is.

Note that $Y$ is not $T_B$ (as long as $Y$ is non-empty and not a singleton).

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  • $\begingroup$ @ Theoretical Economist,,,,,im not getting,.line,......let τY be the trivial topology consisting of only Y and the empty set. f is (trivially) continuous.,,,,,can u elaborate more ? $\endgroup$ – user476275 Oct 26 '17 at 16:30
  • $\begingroup$ The trivial topology is defined as I described it. Any function whose codomain has the trivial topology is continuous. Proving this is an easy exercise, so I will leave that to you. $\endgroup$ – Theoretical Economist Oct 26 '17 at 16:42
  • $\begingroup$ @HennoBrandsma you’re right; I had forgotten about this. I will fix my answer. $\endgroup$ – Theoretical Economist Oct 26 '17 at 21:37

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