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Added compactness tag; http://meta.math.stackexchange.com/questions/4399/would-tags-such-as-ultrafilters-or-stone-cech-compactification-be-too-specif
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Martin Sleziak
Martin Sleziak
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Improved LaTeX.
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Brian M. Scott
Brian M. Scott
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Let $X$ be a compact metric space, show that a continuous function $f:X\rightarrow\mathbb{R}$ attains a maximum and a minimum value on $X$.

Attempt: So the important thing is that I have previously shown that such a function is bounded and that for compact $X$, $f(X)$ is compact given $f$ continuous. In $\mathbb{R}$, compact $\implies$ closed and bounded. So $f(X)$ is closed and contains its accumulation points, and it is bounded so $\exists sup(A),inf(A)$$\exists \sup(A),\inf(A)$ and since closed $\implies sup(A)\in A, inf(A)\in A$$\implies \sup(A)\in A, \inf(A)\in A$.

Did I miss anything/make an unwarranted leap of logic?

Let $X$ be a compact metric space, show that a continuous function $f:X\rightarrow\mathbb{R}$ attains a maximum and a minimum value on $X$.

Attempt: So the important thing is that I have previously shown that such a function is bounded and that for compact $X$, $f(X)$ is compact given $f$ continuous. In $\mathbb{R}$, compact $\implies$ closed and bounded. So $f(X)$ is closed and contains its accumulation points, and it is bounded so $\exists sup(A),inf(A)$ and since closed $\implies sup(A)\in A, inf(A)\in A$.

Did I miss anything/make an unwarranted leap of logic?

Let $X$ be a compact metric space, show that a continuous function $f:X\rightarrow\mathbb{R}$ attains a maximum and a minimum value on $X$.

Attempt: So the important thing is that I have previously shown that such a function is bounded and that for compact $X$, $f(X)$ is compact given $f$ continuous. In $\mathbb{R}$, compact $\implies$ closed and bounded. So $f(X)$ is closed and contains its accumulation points, and it is bounded so $\exists \sup(A),\inf(A)$ and since closed $\implies \sup(A)\in A, \inf(A)\in A$.

Did I miss anything/make an unwarranted leap of logic?

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Emir
Emir
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$X$ compact metric space, $f:X\rightarrow\mathbb{R}$ continuous attains max/min

Let $X$ be a compact metric space, show that a continuous function $f:X\rightarrow\mathbb{R}$ attains a maximum and a minimum value on $X$.

Attempt: So the important thing is that I have previously shown that such a function is bounded and that for compact $X$, $f(X)$ is compact given $f$ continuous. In $\mathbb{R}$, compact $\implies$ closed and bounded. So $f(X)$ is closed and contains its accumulation points, and it is bounded so $\exists sup(A),inf(A)$ and since closed $\implies sup(A)\in A, inf(A)\in A$.

Did I miss anything/make an unwarranted leap of logic?