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enter image description here"Let $X$ be a compact metric space and K(X) is the set of closed subsets of X. That is, $(K(X),d_H)$ is also compact metric space. where $d_H$ is hausdorff metric. If X is finite set, then $[0,1]\times X\to K(X)$ is not continuous. "

: I would like to find an example of this problem. please help me.

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    How do you define the map $[0,1] \times X \to K(X)$? – 2017-02-06
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    Let $B:[0,1]\times X\to K(X)$. $B_r(x)\in K(X)$ where $r\in [0,1]$ for example, $B_0(x)=\{x\}, B_1(x)=X$. $B_r(x)$ is closed ball. ( I do not speak English well. I hope you understand.) – 2017-02-06
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    Let me make sure I understand. $B$ takes a number in $[0,1]$ and a point $x$, and gives back the closed ball of radius that number centered at $x$? – 2017-02-06
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    yes, yes right. – 2017-02-06
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    The image is not about the finite case, though – 2017-02-06

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If $X$ is finite, so is $K(X)$ and $[0,1] \times X$ is just a finite union of compact intervals, each of which must be mapped to a single point of $K(X)$ under $f$ if we have a conrinuous map $f : [0,1] \times X \rightarrow K(X)$. This holds as $K(X)$ is finite $T_1$ so totally disconnected, while $[0,1]$ is connected.

You are considering, per the comment, $f(r,x) = B(r,x)$, the (closed, I suppose) ball of $x$ with radius $r$. So for every fixed $x$: $f[[0,1] \times \{x\}]$ should be constant if $f$ were continuous, but in fact $f(0,x) = \{x\}$ and $f(1,x) = X$ and $X \neq \{x\}$. So for finite but not singleton $X$ the map cannot be continuous.

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    thank you. but i don't understand last "$f(0,x)=\{x\}$ and $f(0,1)=X$ and $X\neq \{x\}$" ... I would appreciate it if you could explain it in more detail. – 2017-02-06
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    I meant $f(1, x)$ and I assume the metric is bounded by $1$. Both are in the same $[0,1]\times \{x\}$ but have different images, contrary what continuity would force. – 2017-02-06
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    thank you so much. It is a great help to study. thank you!! :) – 2017-02-06
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    @hyunheelee glad I could help. – 2017-02-06
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    If you okay, I just want to ask you one more question. Can you explain the picture above? please click "enter images". – 2017-02-06