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Let $X, Y$ be topological spaces, with Y a Hausdorff space. Prove that if there exists an injective and continuous function $f: X \rightarrow Y$, then $X$ is Hausdorff.

Here's my idea:

Since $Y$ is Hausdorff, different points $x,y \in Y$ have disjoint neighborhood $U \subset \tau_x,V \subset \tau_y$.

But $f$ is also continuous, so the pre-image of these sets is an open set in $X$, say $f^{-1}(U)=A$, $f^{-1}(V)=B$. Now, because $f$ is injective, I got that $A \cap B = \emptyset$, and $f^{-1}(x)$, $f^{-1}(y)$ have disjoint neighborhoods.

José Carlos Santos
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VoB
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4 Answers4

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Let $x_1,x_2 \in X$ such that $x_1 \neq x_2$

Then $f(x_1),f(x_2) \in Y$ and $f(x_1) \neq f(x_2)$ because $f$ is injective.

Then there exist two disjoint open sets $U_1,U_2 \in \mathcal{T_Y}$ such that $f(x_1) \in U_1$ and $f(x_2) \in U_2$

Thus $$x_1 \in f^{-1}(U_1)$$ $$x_2 \in f^{-1}(U_2)$$ where $ f^{-1}(U_1),f^{-1}(U_2)$ are open in $X$ because $f$ is continuous and $U_1,U_2$ are open in $Y$.

Put $V_1=f^{-1}(U_1)$ and $V_2=f^{-1}(U_2)$ and you have: $$V_1 \cap V_2 =f^{-1}(U_1) \cap f^{-1}(U_2)=f^{-1}(U_1 \cap U_2)=f^{-1}(\emptyset)=\emptyset$$

Therefore $X$ is Hausdorf.

Marios Gretsas
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There's a jump in your proof: you write that $x$ and $y$ have disjoint neighborhoods $U\in\tau_x$ and $V\in\tau_y$ (note that you wrote $U\subset\tau_x$ and $V\subset\tau_y$; this is wrong), and, in the next sentence, you write that the pre-images of $U$ and $V$ are open sets. Why?

There are two ways of solving this:

  1. Use the fact that the pre-image of a neighborhood by a continuous function is again a neighborhood.
  2. You can assume without loss of generality that $U$ and $V$ are open sets.
José Carlos Santos
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  • I said that $U, V \subset Y$ are open sets just because $f$ is continuous. – VoB Aug 22 '17 at 22:14
  • The existence of such neighborhoods have nothing to do with $f$. In my opinion, you should have started with two neighboroods $U^\star$ and $V^\star$ of $f(x)$ and of $f(y)$ respectively such that $U^\star\cap V^\star=\emptyset$ and then say that there are open sets $U\in\tau_x$ and $V\in\tau_y$ such that $x\in U\subset U^\star$ and $y\in V\subset V^\star$. Since $U^\star\cap V^\star=\emptyset$, $U\cap V=\emptyset$. – José Carlos Santos Aug 22 '17 at 22:20
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Take distinct $x,y \in X$ then $f(x), f(y)$ are distinct points in $Y$ (by injectivity) and so we have disjoint open sets $U$ and $V$ containing $f(x)$ and $f(y)$, respectively. Now their preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open (by continuity), disjoint and contain $x$ and $y$.

Zain Patel
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Let $x_1,x_2$ be distinct in $Y$. By injectivity of $f$, $f(x_1),f(x_2)$ are distinct. Since $Y$ is hausdorff, there exists disjoint neighborhoods $U_i$ of $f(x_i)$. By continuity of $f$, $f^{-1}(U_i)$ is open for each $i$ and contains $x_i$. Since $f^{-1}(U_1)\cap f^{-1}(U_2)=f^{-1}(U_1\cap U_2)=f^{-1}(\varnothing)=\varnothing$, $X$ is hausdorff.