Continuous Mapping from Compact Space to Hausdorff Space Preserves Local Connectedness

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Theorem

Let $T_1 = \left({S_1, \tau_1}\right)$ be a compact topological space.

Let $T_2 = \left({S_2, \tau_2}\right)$ be a $T_2$ (Hausdorff) space.

Let $f: T_1 \to T_2$ be a continuous mapping.

Let $T_1$ be locally connected.


Then $T_2$ is also locally connected.


Proof

Let $H$ be a component of an open set $U$ of $T_2$.

By definition of continuous mapping, $f^{-1} \left[{U}\right]$ is an open set of $T_1$.

Let $G$ be a component of $f^{-1} \left[{U}\right]$.

Thus by Continuous Image of Connected Space is Connected, $f \left[{G}\right]$ is connected in $T_2$.

Thus either:

$f \left[{G}\right] \subseteq H$

or:

$f \left[{G}\right] \cap H = \varnothing$

Thus every component of $f^{-1} \left[{H}\right]$ is a component of $f^{-1} \left[{U}\right]$.


From Component of Locally Connected Space is Open, $f^{-1} \left[{H}\right]$ is open.

Let $S_1 \setminus G$ denote the complement of $f^{-1} \left[{H}\right]$ relative to $S_1$.

By definition of closed set, $S_1 \setminus f^{-1} \left[{H}\right]$ is closed in $T_1$.

From Closed Subspace of Compact Space is Compact, $S_1 \setminus f^{-1} \left[{H}\right]$ is also compact in $T_1$.

From Continuous Image of Compact Space is Compact, $f \left[{S_1 \setminus f^{-1} \left[{H}\right]}\right] = S_2 \setminus H$ is compact in $T_2$.

From Compact Subspace of Hausdorff Space is Closed, $S_2 \setminus H$ is closed in $T_2$.

Thus by definition of closed set, $H$ is open in $T_2$.


$\blacksquare$


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