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There's a problem in my text which reads as:

Show that $C[0,1]$ is not an open subset of $(B[0,1],\|.\|_\infty).$

I've already shown in a previous example that for any open subspace $Y$ of a normed linear space $(X,\|.\|),~Y=X.$ Even though using this result this problem turns out to be immediate the sup-norm is becoming immaterial.

And I can't believe what I'm left with:

No norm on $B[0,1]$ can be found to make $C[0,1]$ open in it.


Is this a correct observation?

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  • $\begingroup$ Why can you not believe it? $\endgroup$ – Chris Eagle Jul 13 '13 at 13:17
  • $\begingroup$ Because so simply I get to conclude such a huge statement. $\endgroup$ – Sriti Mallick Jul 13 '13 at 13:18
  • $\begingroup$ It's not a huge statement. $\endgroup$ – Chris Eagle Jul 13 '13 at 13:19
  • $\begingroup$ I meant to say I didn't expect to predict about all possible norms so easily. $\endgroup$ – Sriti Mallick Jul 13 '13 at 13:21
  • $\begingroup$ But any theorem about normed spaces is a statement about all possible norms. Do you have this reaction whenever you read a theorem? $\endgroup$ – Chris Eagle Jul 13 '13 at 13:26
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This statement is actually true under more general settings. It seems convenient to talk about topological vector spaces, of which normed spaces are a very special kind.

So let $X$ be a topological vector space and $Y$ be an open subspace. So we know $Y$ contains some open set.

Since the topology on topological vector spaces are translation invariant (that is, $V$ is open if and only if $V+x$ is open for all $x\in X$, you can check this in normed spaces), we know $Y$ contains some open neighborhood of the origin, say $0\in V\subset Y$.

Another interesting fact about topological vector spaces is that for any open neighborhood $W$ of the origin, one has \begin{equation} X=\cup_{n=1}^{\infty}nW. \end{equation}Again you might check this for normed spaces. Apply this to our $V$, and note that $Y$ is closed under scalar multiplication, we have \begin{equation} X=\cup nV\subset \cup nY=Y. \end{equation}

So we have just proved

The only open subspace is the entire space.

Note: If $S$ is a subset of a vector space, for a point $x$ and a scalar $\alpha$ we define \begin{equation} x+S:=\{x+s|s\in S\} \end{equation} and \begin{equation} \alpha S:=\{\alpha s|s\in S\}. \end{equation}

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  • $\begingroup$ Your "interesting fact about topological vector spaces" seems to make some assumptions about the underlying field. What are they exactly? $\endgroup$ – Niels J. Diepeveen Jul 13 '13 at 14:19
  • $\begingroup$ @NielsDiepeveen I assume we are talking about vector spaces over $\mathbb{R}$ or $\mathbb{C}$. $\endgroup$ – Hui Yu Jul 13 '13 at 14:22
  • $\begingroup$ I do not know any topological vector spaces built on fields other that $\mathbb{R}$ or $\mathbb{C}$. $\endgroup$ – Hui Yu Jul 13 '13 at 14:50
  • $\begingroup$ @NielsDiepeveen Do we really want to do functional analysis on discrete fields? $\endgroup$ – Hui Yu Jul 14 '13 at 0:54
  • $\begingroup$ @NielsDiepeveen What is the use of topological vector spaces if one is not doing functional analysis? XD $\endgroup$ – Hui Yu Jul 14 '13 at 11:35
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This is true. Indeed, you can show more:

Let $X$ be a normed vector space. If $E \subseteq X$ is a linear subspace which is open, then $E = X$.

My guess is this follows from the same argument you already have.

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    $\begingroup$ Er, the OP explicitly states this is the argument they used. $\endgroup$ – Chris Eagle Jul 13 '13 at 13:18
  • $\begingroup$ @ChrisEagle: Oh, I missed that sentence. $\endgroup$ – Nate Eldredge Jul 13 '13 at 13:28
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    $\begingroup$ If you want to generalize: a proper subgroup of a connected topological group has empty interior. That's still very easy. $\endgroup$ – Niels J. Diepeveen Jul 13 '13 at 13:43

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